Usually **scientists** will collide a large** isotope** such as** Uranium-235** with a **neutron**. The collision results in several more nuclear fissions, also known as a chain reaction . This results in a HUGE release in energy!

**Nuclear fission **occurs when** large isotopes** such as uranium - 235 collide with neutron and split into two or more **smaller atoms. **and releases a large quantity of energy. in this process nucleus of atom split into two or more smaller nuclei . this result in huge release in energy and resulting in chain reaction.

Thus, Usually **scientists** will collide a large** isotope** such as** Uranium-235** with a **neutron**. The collision results in several more nuclear fissions, also known as a chain reaction . This results in a HUGE release in energy!

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metric preferred to write 5.8 * 10 to -3 g

Since the number is 5.8*10^-3 g, the negative symbol in front of the power of 3 means that there are 3 0's in front of the real number, in standard notation, therefore we will add these 3 zeros to the number

0.0058 grams, and if we multiply it by 1000 we will end up with 5.8 mg, letter C

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. 2CH3OH(g)+3O2(g)⟶2CO2(g)+4H2O(g)

The **enthalpy **of the **reaction **is obtained as -430.22 kJ/mol

The** enthalpy change** refers to the energy that is **lost **or **gained **in the reaction. Let us now look at the standard enthalpies of formation of the reactants and the products.

Standard enthalpy of formation of gaseous water = -241.82 kJ/mol

Standard enthalpy of formation of gaseous carbon dioxide = -393.5 kJ/mol

Standard enthalpy of formation of gaseous methanol = -205.10 kJ/mol

Standard enthalpy of formation of gaseous oxygen = 0 kJ/mol

Hence;

Using the formula;

Enthalpy of formation of the products - Enthalpy of formation of the reactants

ΔH = [(-393.5 ) + (-241.82)] - [(-205.10) + 0]

ΔH = (-635.32) + 205.10

ΔH = -430.22 kJ/mol

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a graduated cylinder contains 25.8 ml of water.Due drops 2 glass marbles weighting 6.5 g into the cylinder.Thew new water level reads 27.2 ml what is the volume of the marbles.What is the density of the marbles

The **volume** of the 6.5g **marble** if it was dropped into a graduated cylinder containing 25.8 ml of water is **1.4mL**.The **density** of the **marble** is 4.64g/mL. How to calculate volume?

According to this question, a graduated cylinder contains 25.8 mL of water. 2 glass marbles weighing 6.5 g was dropped into the cylinder. The new water level reads 27.2 mL.

This means that the **volume** of the **marbles** can be calculated as follows:

27.2mL - 25.8mL = 1.4mL

The **volume** of the **marbles** is 1.4mL.

**Density** is a measure of the mass of matter contained by a unit volume. The **density** can be calculated by dividing the mass by the volume as follows:

**Density** = 6.5g ÷ 1.4mL = 4.64g/mL

The **density** of the **marble** is 4.64g/mL.

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Name something you use in your home that could be considered an acidic, basic, and neutral solution. Use properties of acids and bases to tell why do you think the solutions you chose could be considered acidic, basic, or neutral.

Acidic solution I use: Lemon Juice (Citric acid)

I know this is **acidic** because it is **sour** in taste just like all other acids.

Basic solution I use: Handwash

I know this is **basic **because it is **soapy** and **slippery **in nature, and it **dissolves oil **and grease.

Neutral solution I use: Water

I know this is **neutral** because it has **equal **number of H+ and OH- ions.

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Arrange the following elements in order of decreasing atomic size.

Rank the atoms from the largest to smallest. To rank items as equivalent, overlap them

The arrangement of the in order of decreasing** atomic size** is as follows:

The **atomic size** of an atom is the diameter of the atom and is calculated as half the distance between the nuclei of two covalently-bonded nuclei of the atom.

The **atomic size** of elements increases down a group but decreases across a period.

The reason for this trend is that, on going down a group, an extra electron shell is added to the element below. However, across a period, the electron shell remains the same whereas the size of the positive nuclear charge increases, therefore, the atomic size of atoms in the same period decreases from left to right across the period.

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Complete question:

Arrange the following elements in order of decreasing atomic size.

Rank the atoms from the largest to smallest. To rank items as equivalent, overlap them: Al, Cl, Ar, P, and K.

___Mg(s) + ___O2(g) ____Mg(s)

The balanced **equation** for the reaction above is: **2**Mg(s) + O₂(g) = **2**MgO(s).

To balance a chemical reaction or equation, it simply means to make sure the total number of **atoms** of elements on the reactants side of the equation **equals** or is the same as the number of atoms in product side.

From the task given above, to balance the equation in the reaction between **magnesium** and **oxygen** to give the product, magnesium oxide:

Mg(s) + O₂(g) = MgO(s)

Since there are 2 oxygen atoms in reaction side and 1 oxygen atom at the product side, you'll add 2 to the product **(MgO)** so we can have the same number of oxygen atoms as the reactant side and complete it by adding 2 also to **magnesium** in the product side:

2Mg(s) + O₂(g) = 2MgO(s

So therefore, having the same atom numbers in both reactants and products side is the only way to balance chemical equation.

**Complete question:**

Balance the chemical reaction below:

___Mg(s) + ___O₂(g) ____MgO(s)?

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Calculate the number of grams of glucose (C6H12O6) that would need to be dissolved to a total volume of 1.5L to get a 15.3% (w/v) solution

**ANSWER**

The mass of glucose in grams is **230 grams**

**EXPLANATION**

Given information

The total volume of the solution = **1.5L**

Follow the steps below to find the mass of glucose

Step 1: Convert the volume of the solution from L to mL

According to the standard conversion, 1L is equivalent to 1000mL

Let x represents the volume of the solution in mL

[tex]\begin{gathered} \text{ 1L }\rightarrow\text{ 1000mL} \\ \text{ 1.5L }\rightarrow\text{ xmL} \\ \text{ Cross multiply} \\ \text{ 1L }\times\text{ xmL = 1000mL }\times\text{ 1.5L} \\ \text{ Isolate x} \\ \text{ xmL = }\frac{1000mL\times\text{ 1.5L}}{1L} \\ \text{ x = }\frac{1000\times1.5\cancel{L}}{1\cancel{L}} \\ \text{ x = 1500mL} \end{gathered}[/tex] Hence, the volume of the solution in mL is **1500mL**

Step 2: Find the mass of the glucose in grams

[tex]\begin{gathered} \text{ The mass of glucose = }\frac{15.3}{100}\times\text{ 1500} \\ \text{ The mass of glucose = 0.153 }\times\text{ 1500} \\ \text{ The mass of glucose = 229.5 grams} \end{gathered}[/tex]Hence, the mass of glucose in grams is **229.5 grams**

Determine the mass in grams of HCl that can react with 0.750 g of Al(OH)₃ according to the following reaction

Al(OH)₃(s) + 3 HCl(aq) → AlCl₃(aq) + 3 H₂O(aq)

How do i start this? I need to convert from moles to grams of HCI but not sure where to start

1.05g of HCl can react with 0.750g of Al(OH)3 to produce the products in the equation.

Corn oil, ice, and ethanol are mixed in the same container. What will be the order of the liquids once they settle out from top to bottom ? A. corn oil, ice, ethanol B. ethanol, ice, corn oil C. ice, ethanol, corn oil D. corn oil, ethanol, ice

The order of the liquids will depend on the density of each of them. The densest will go to the bottom of the container, the least dense will be on top.

Let's see what the density of each of the substances is, let's assume that we are at room temperature.

Ice density = 0.917g/cm^3

Corn oil density =0.925g/cm^3

Ethanol density = 0.789g/cm^3

The order of density from lowest to highest is:

Ethanol < Ice < Corn oil

This will be the order from top to bottom, so the answer will be:

**B. ethanol, ice, corn oil C**

Which has the incorrect name-formula combination? A) Cobalt(II) chlorite - Co(CIO2)2 B) Iron(II) chlorate - FECIO4 C) Manganese(II) perchlorate - Mn(CIO4)2 D) Chromium(III) hypochlorite - Cr(CIO)3

The incorrect name-**formula** combination is FeCIO₄ - Iron(I) perchlorate; option B.

The** IUPAC nomenclature** of naming compounds refers to the convention of naming **compounds** using the oxidation states of the elements present in the **compound**.

The **IUPAC** **nomenclature** of the given compounds is given below as follows:

A. Co(CIO₂)₂ - Cobalt(II) chlorite; the oxidation state of cobalt is +2

B. FeCIO₄ - Iron(I) perchlorate; the **formula** of the compound is wrong because iron does not exist in the +1 state

C)Mn(CIO4)₂ - Manganese(II) perchlorate; the oxidation state of manganes is +2

D) Cr(CIO)₃ - Chromium(III) hypochlorite ; the oxidation state of chromium is +3

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Predict the missing component

in the nuclear equation.

175/71 Lu → 4/2 He + X

**Answer:**

B

**Explanation:**

What is the amount of valence electrons in an atom equivalent to?The amount of protons an atom hasThe group the atom is locatedThe period the atom is locatedThe amount of electrons an atom has

ANSWER

The amount of valence electron in an atom is equivalent to **the group the atom is located**

**EXPLANATION**

Valency is define as the combining power of an element. Valence electrons are electrons located at the outer most shell of an element.

Valence electrons tell us the group a particular element belongs to and also the chemical properties of the element.

For example, sodium atom

Sodium has the below electronic configuration

[tex]\text{ 1s}^2\text{ 2s}^2\text{ 2p}^6\text{ 3s}^1[/tex]The structure above represents a sodium atom. The structure contains a single electron at the outer most shell of the atom. The single electron is called the valence electron of sodium. Hence, sodium belongs to group 1

Therefore, the amount of valence electron in an atom is equivalent to **the group the atom is located**

What is the mass percentage of carbon in 5.00g of sucrose? 50.00g of sucrose? 500.0g of sucrose?

Because amount has no effect on **mass **percentages, the **mass **percentage of **carbon **in 5, 50, and 500 g of sucrose is** 42.1 %**

**The mass:**

It also represents the body's inertia, or resistance to **acceleration** (change in velocity) when a** net force** is applied. The mass of an object dictates the strength of its gravitational attraction to other** bodies.**

The** kilogram** is the SI base unit of mass (**kg**). **Mass **is not the same as weight in physics, despite the fact that** mass **is frequently calculated by **measuring** the object's weight using a spring scale rather than a balance scale and comparing it directly with known **masses**. Because of the reduced** gravity**, an object on the Moon would weigh less than it does on** Earth,** yet it would still have the same **mass**.

**Molecular **formula of sucrose is C12 H22 O11

**Molar mass **of Sucrose , C12 H22 O11 is = ( 12 x 12 ) + ( 22 x 1 ) + ( 11 x 16 ) g/mol

= 342 g/mol

The **molar mass **of Carbon is 12 g/mol

342 g of C12 H22 O11 contains 12 x 12 g of C

5 g of C12 H22 O11 contains Y g of C

Y = ( 12x12x 5 ) / 342

= 2.105 g of C

mass % of C = (mass of C / mass of sucrose ) x 100

= (2.105 / 5 ) x 100

= 42.1 %

342 g of C12 H22 O11 contains 12 x 12 g of C

50.0 g of C12 H22 O11 contains Y g of C

Y = ( 12x12x 50 ) / 342

= 21.05 g of C

mass % of C = (mass of C / mass of sucrose ) x 100

= (21.05 / 50 ) x 100

**= 42.1 %**

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What is the pH of a 1.63 M solution of methylamine?

Report your answer to 2 decimal places.

The **pH** of the 1.63 M solution of methylamine is 10.56.

What is the pH of the solution?

The **pH** of the 1.63 M solution of **methylamine** is calculated from the equation of the dissociation reaction as follows:

H₃CNH₂ (aq) + H₂O (l) ---> H₃CNH₃ (aq) + OH⁻ (aq)

pKb of H₃CNH₂ = 3.66

Kb of H₃CNH₂ = 10⁻³°⁶⁶ = 2.19 * 10⁻⁴

The [OH¯] is calculated using the Kb expression:

Kb = [OH⁻] * [H⁺] / [HA}

[OH⁻] = [H⁺] = x

Kb = x² / (1.63 - x)

Assuming is x <<< 1.63, x is negligible and the denominator = 1.63

x = √(2.19 * 10⁻⁴ x 10⁻¹¹ * 1.63)

x = 3.57 * 10⁻⁴

pOH = -log (3.57 * 10⁻⁴)

pOH = 3.44

pH = 14 - 3.44

pH = 10.56

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Now that you have learned to balance equations using the models on the simulation. See if you can balance the(These are not found in the sim). Balance the reactions given

1)

Na3PO4 + H2SO4 => Na2SO4 + H3PO4

We must have the same number of atoms on both sides.

We start with Na,

2 Na3PO4 + H2SO4 =>3 Na2SO4 + H3PO4 (6 Na on both sides)

Then we proceed with P,

2 Na3PO4 + H2SO4 =>3 Na2SO4 +2 H3PO4 (2 P on both sides)

After that, we continue with S,

2 Na3PO4 + 3 H2SO4 =>3 Na2SO4 +2 H3PO4 (3 S)

The number of H and O are solved.

**Answer: 2 Na3PO4 + 3 H2SO4 =>3 Na2SO4 +2 H3PO4**

In its native state, which element has bonds between many cations and a sea of valence electrons?

O He

O CI

O Zn

Ο Ν

**Answer:** The correct answer is **C. Zn**

**Explanation:** Zn in its native state has bonds between multiple cations and a sea of valance electrons.

Calculate the molarity of a carbonic acid solution given the following titration results: 47.00 mL of the carbonic acid solution was neutralized to a phenolphthalein endpoint with 23.82 mL of 0.1250 M ammonium hydroxide.

To know the molarity of carbonic acid when titrated with ammonium hydroxide. We use [tex]M_{1} V_{1} =M_{2} V_{2}[/tex]** formula** and hence the molarity of carbonic acid is ** 0.063M.**

**Titration** is an experimental technique in which the **molarity** of unknown solution is calculated using other solution whose molarity is known. To know the end point we use phenolphthalein as** indicator. **

Mathematically,

[tex]M_{1} V_{1} =M_{2} V_{2}[/tex]

where,

[tex]M_{1}[/tex]=Molarity of carbonic acid

[tex]M_{2}[/tex]=Molarity of ammonium hydroxide

[tex]V_{1}[/tex]=Volume of carbonic acid

[tex]V_{2}[/tex]=Volume of ammonium hydroxide

Substituting all values

[tex]M_{1}[/tex]=(0.125×23.8)÷47.00

[tex]M_{1}[/tex]=0.063M

Thus the** molarity** of carbonic acid is **0.063M**

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Why D is the correct answer

**Explanation:**

Because during the phase transformation the pressure inside the closed container remains the same and that's a fact ay

how could armondo test how curves in a river affect the speed of the river

Armondo could test how curves in a **river **affect the speed of the river as the **speed **of the water outside of a bend increases as a river rounds the bend. The water's **velocity**, however, diminishes as it approaches the interior of the curve.

A fantastic illustration of how water can alter the contour of the land is a meandering **river**. A river seldom turns when it is bordered by sheer rock, but it will do so when it opens up in broad **valleys**. Water will flow more swiftly and destroy the ground faster outside the river. Over time, it will **curve **too much and slow down.

To simulate a meandering river, use a stream table. The **speed **of the water outside of a bend increases as a river rounds the bend. The water's **velocity**, however, diminishes as it approaches the interior of the curve. A bar of deposited silt, like this one, is created as a result of the reduction in velocity.

The velocity of a river is the rate at which water flows through its course. Numerous elements, such as the **channel's **form, the **slope's **grade, the amount of water carried by the river, and the amount of friction brought on by jagged edges in the **riverbed**, all affect a river's speed.

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An excess of chromium metal is added to 500.0 mL of a 0.915 M AgNO3solution in a constant-pressure calorimeter. As a result of the reactionCr(s) + 2 AgNO3(aq)Cr(NO3)(aq) + 2 Ag(s)the temperature rises from 19.3 °C to 55.9 °C. Based on your previoustwo answers, calculate reaction (in J).Please help I don’t understand how i got it wrong :(

The **enthalpy **of the **reaction **is -164 kJ/mol.

We know that the reaction that occurs between the **chromium metal** and the **acid **is an **exothermic **reaction thus there is an increase in the temperature of the system.

Number of moles of the silver nitrate solution is obtained from;

Volume * concentration

500/1000 L * 0.915 M = 0.46 moles

We can now assume that the density of the solution is 1 g/mL hence the mass of the solution is 500g. Let the specific heat capacity of the solution be 4.18 J/Kg/°C.

Then;

H = mcdT

H = Heat lost in the reaction

m = mass of the solution

c = specific heat capacity

dT = temperature change

H = 500 * 4.12 * ( 55.9 - 19.3)

= 75.4 kJ

The heat of reaction = 75.4 kJ/0.46 moles

= -164 kJ/mol

Let us recall that the negative simply means that heat was lost in the reaction.

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For the reaction: 2 A (g) + B (s) ⇌ 2 C (s) + D (g) Kp = 8210

At 298 K in a 10.0 L vessel, the known equilibrium values are as follows: 0.021 atm of A, 0.22 mol of B, and 10.5 mol of C. What is the equilibrium partial pressure of D?

**Answer:D**

**Explanation:**

The equilibrium **partial pressure **of D is 0.362atm.

**2A(g) + B(s) <==> 2C(s) + D(g)**

The pressure that a gas, in a **mixture** of gases, would exert if it alone occupied the whole volume occupied by the mixture is called **partial pressure** of the gas.

In the equilibrium expression, we ignore **solids** as they are considered to have a value of unity. Thus, we can write the equilibrium expression for the above reaction as:

**K = (PD) / (PA)² **

where, K = Partial pressure equilibrium

PD = Partial pressure of gas D

PA = Partial pressure of gas A

Kp = 8210

PA = 0.021 atm

8210 = PD / (0.021)²

PD = 8210 × (0.021)²

PD = 0.362 atm.

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The specific heat of a certain type of cooking oil is 1.75 J/(g-°C). How much heat energy is needed to raise the temperature of 2.74 kg of this oil from 23 °C to 191 °C?

The amount of **heat energy** needed to increase a substance's temperature by 1°C per unit mass is known as its** specific heat capacity** (or 1K). joule per kilogram per kelvin is the SI unit (Jkg-1K-1).

so,

805,560 Joules** heat energy** is needed to raise the temperature from 23 °C to 191 °C.

Use the formula:

Q = cMΔT

where

Q = **heat energy** needed for that material to get desired temperature change (in Joules)

M = mass (in grams) so you have to convert from kilograms.

c = **specific heat** constant for the material being heated [in /(grams oC)]

ΔT = change in temperature (in oC)

Q = (1.75)(2.74 x 1000)(191 - 23) = (1.75)(2740)(168) = 805,560 Joules

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**Q = cMΔT**

where

Q is the amount of heat energy required to change the material's temperature.

You must convert from kilograms since M is mass (in grams).

c = the material's specific heat constant [in grams per degree Celsius]

T stands for temperature change (in oC)

Q is equal to (1.75 x 2.17 x 1000) (191 - 23) = (1.75 x 2170 x 168) = **637,980** **Joules.**

**What does "specific heat capacity" mean?**

In **thermodynamics**, a substance's specific heat capacity, commonly referred to as massic heat capacity, is calculated by dividing its heat capacity by its mass in a sample.

particular heat capacity's mathematical formula?

**Delta T = Q=mc**

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Scientists discover a species that is unicellular and has chloroplasts. What could it be

**Answer: dinophyte**

**Explanation: its a type of protist. it is unicellular and has chloroplasts**

Scientists discover a species that is unicellular and has chloroplasts. It could be a **dinophyte. **

A **single-celled **eukaryotic group (Protists) called a dinophyte makes up the phylum **Dinoflagellates. **They possess **chloroplasts **and are single-celled.

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If a solution containing 24.68 g of mercury(II) nitrate is allowed to react completely with a solution containing 7.410 g of sodium dichromate, How many grams of the reactant in excess will remain after the reaction?

The balanced equation is

[tex]Hg(NO_3)_2+Na_2Cr_2O_7\to HgCr_2O_7+2Na(NO_3)[/tex]We need the molar mass of each compound.

• The molar mass of Hg(NO3)2 is 324.7 g/mol.

,• The molar mass of Na2Cr2O7 is 261.97 g/mol.

,• The molar mass of HgCr2O7 is 416.58 g/mol.

,• The molar mass of 2Na(NO3) is 169.99 g/mol.

Then, we find the number of moles of each reactant.

• Moles of Hg(NO3)2 = 24.68g ÷ 324.7 g/mol = 0.076 mol.

,• Moles of Na2Cr2O7 = 7.41g ÷ 261.97 g/mol = 0.028 mol.

According to equation 1 mole of Hg(NO3)2 will react with 1 mole of Na2Cr2O7, which means 0.076 mol of Hg(NO3)2 has to react with 0.076 mol Na2Cr2O7, which is not possible because there are not enough moles to get 0.076 of Na2Cr2O7 in the reaction. Hence, Na2Cr2O7 is the limiting reactant and Hg(NO3)2 is the excess reactant.

Then, subtract the number of moles to obtain the excess:

The remaining moles from Hg(NO3)2 are: 0.076mol - 0.028 = 0.048 mol.

**Therefore, the remaining excess reactant is 0.048 moles.**

But, we need to transform it to grams using the molar mass of Hg(NO3)2.

[tex]0.048\text{mol}\cdot\frac{324.7g}{1\text{mol}}=15.59g[/tex]**Therefore, the remaining mass of Hg(NO3)2. is 15.59 grams.**

If 2.47 g of CuNO3 is dissolved in water to make a 0.820 M solution, what is the volume of the solution in milliliters?

volume:

**Answer:**

23.4 milliliters

**Explanation:**

Note 1: This answer assumes that the volume of CuNO3 is negligible

Note 2: CuNO3 can't be produced in any meaningful quantities and can't be obtained by the average chemist, maybe you meant Cu(NO3)2 instead?

From the definition of molarity, molarity = moles / volume

the number of moles is the number of grams divided by the molar mass, or

2.47 divided by 125.55 which is 0.01967

The M and moles is known so volume can be found.

0.82 = 0.01967 / volume

0.82 * volume = 0.01967

volume = 0.01967 / 0.82 = 0.023988 liters = 23.4 milliliters

Oxidation number of each element in the compound (NH4)2CrO4

**Answer: **

**the oxidation number of N in (NH4)2CrO4 is -3. the oxidation number of O in (NH4)2CrO4 is -2. the oxidation number of Cr in (NH4)2CrO4 is +6. the oxidation number of H in (NH4)2CrO4 is +1**

**hope i helped <3**

N has an **oxidation number **of -3, O has an oxidation number of -2, and Cr has an oxidation number of +6 in the compound **(NH4)2CrO4.** The H in (NH4)2CrO4 has an oxidation number of 1

**Oxidation number **is defined as how many **electrons **a given atom or ion has either **gained **or **lost **when compared to a **neutral atom**. Any free element's oxidation number is always zero. The meaning was expanded to cover additional reactions in which **electrons **are **lost**, regardless of whether oxygen was present, since the substance loses electrons when it is oxidized.

In the case of a **monatomic ion**, the oxidation number is always the same as the ion's charge. The name of the compound is **ammonium chromate **which has oxidation number both in positive as well as negative charge electrons.

Thus, N has an **oxidation number **of -3, O has an oxidation number of -2, and Cr has an oxidation number of +6 in the compound **(NH4)2CrO4.** The H in (NH4)2CrO4 has an oxidation number of 1.

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Predict the products for each of the following reactions. Write the molecular equation, the complete ionic equation, and the net ionic equation. Classify the reactions in as many ways as you can.

a. Aqueous copper (II) chloride is added to aqueous silver nitrate.

The molecular equation of the reaction of** copper chloride** and aqueous **silver nitrate**:

CuCl₂ (aq) + 2 AgNO₃ (aq) → Cu(NO₃)₂ (aq) + 2AgCl (s)

The **net ionic equation** will be:

2Ag⁺(aq) + 2Cl⁻(aq) → 2AgCl(s)

What are the net ionic equations?The **net ionic equation** can be described as an equation that represents only those elements, **compounds**, or **ions** that directly participated in that particular chemical reaction.

The** balanced chemical equation **reaction of copper chloride and aqueous silver nitrate:

CuCl₂ (aq) + 2 AgNO₃ (aq) → Cu(NO₃)₂ (aq) + 2AgCl (s)

The** complete ionic equation** for the reaction can be represented as follows:

Cu²⁺(aq) + 2Cl⁻(aq) + 2Ag⁺(aq) + 2NO₃⁻ → Cu²⁺ (aq) + 2NO₃⁻(aq) + 2AgCl(s)

In the ionic equation, the **copper** and **nitrate ions **appear unchanged on both sides of the equation. When we mix the two solutions, the **copper** and nitrate ions do not participate in the reaction. So copper and **nitrate** ions can be **eliminated **from the ionic equation.

2Ag⁺(aq) + 2Cl⁻(aq) → 2AgCl(s)

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How does this simulation demonstrate Newton’s third law of motion?

**Answer:**

For every action, there is opposite and equal reaction

A mixture of KCl and KNO3 is 44.20% potassium by mass. The percent of KCl in the mixture is closest to

Step-by-step explanation:

Assuming a basis of 100 grams

composition of Potassium = 44.2%

composition of potassium = 0.442

mass of potassium = 0.442 x 100g

Mass of potassium= 44.2 grams

mole of potassium = reacting mass / molar mass

Molar mass of potassium = 39.1 gram/ mol

Mole of potassium = 44.2 / 39.1

Mole of potassium = 1.13 moles

mole of potassium chloride + mole of potassium nitrate = mole of potassium

n(KCL) + n(KNO3) = n(k)

since mole (n) = m/M.M

Then we have,

m(KCL)/M.M(KCL) + m(KNO3)/M.M (KNO3) = 1.13

M.M (KCL) = 39.1 + 35.5

M.M (KCL) = 74.6 gram/mol

M.M (KNO3) = 39.1 + 14 + 3(16)

M.M (KNO3) = 39.1 + 14 + 48

M.M (KNO3) = 101.1 gram/mol

Let the mass of KCL be x

Let the mass of KNO3 be y

Assuming the total mass of the mixture is 100g

x + y = 100 ---------- equation 1

x/74.6 + y/101.1 = 1.13 ---- equation 2

From the first equation, make x the subject of the formula

x = 100 - y

100- y / 74.6 + y/101.1 = 1.1.3

1.355(100 - y) + y = 1.13 * 101.1

135.52 - 1.355y + y = 114.243

Collect the like terms

-1.355y - y = 114.243 - 135.52

-0.355y = -21.277

Divide both sides by -0.355

-0.355y/-0.355 = -21.277/-0.355

y = 59.9grams

y is 60 grams approximately

Recall, y is the mass of KNO3

From the first equation

x + y = 100

How many moles are in .009 grams of Carbon?

The no. Of **moles **of carbon is found to be 7.5×10⁻⁴.

No. of moles are give by = Given mass/**molar mass**.

Given mass is 0.009g and the molar mass of carbon atom is 12g.

So, the no. Of moles are given by ,

No. Of moles = 0.009/12

= 0.00075 = 7.5×10⁻⁴mol.

The value of 6.023 x 10²³ is equal to one mole of any substance (**Avogadro number**). It can be used to quantify the chemical reaction’s by-products. The symbol for the unit is mol.

Carbon element is insoluble in water, diluted acids and bases, as well as organic solvents, carbon is an inert material. It combines with oxygen at high temperatures to generate **carbon monoxide** or dioxide. Diamond and graphite are two distinct** allotropes**, crystalline forms of the carbon atom.

To learn more about **allotropes**, refer this link.

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