1. What is WRONG with the setup used by the student to solve this

Before the Internet, it was easier for small businesses to do business on a global scale. Choose the answer.

Umnica [9.8K]

I think it’s false ......

8 0

3 years ago

An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

jekas [21]

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

For example, $O_{2}$ , $N_{2}$ etc are pure substances.

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

pure chemical substance.

heterogenous mixture.

mixture.

4 0

3 years ago

A sample of Mo(NO3)6 has 2.22 x 10^22 nitrogen atom, how many oxygen atoms does the sample have?

goldfiish [28.3K]

There are 3.98 × 10^23 atoms of oxygen in the sample.

Given that;

1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen

x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen

x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms

x = 0.0368 moles

The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18

Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.

Learn more: brainly.com/question/9743981

4 0

2 years ago

A 1.50 g sample of solid NH₄NO₃ was added to 35.0 mL of water in a styrofoam cup (insulated from the environment) and stirred un

GaryK [48]

Answer : The heat of the reaction is, 1.27 kJ/mole

Explanation :

First we have to calculate the heat released.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat = ?

m = mass of sample = 1.50 g

c = specific heat of water = $4.81J/g^oC$

$T_1$ = initial temperature = $22.7^oC$

$T_2$ = final temperature = $19.4^oC$

Now put all the given value in the above formula, we get:

$Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC$

$Q=-23.8095J=-0.0238kJ$

Now we have to calculate the heat of the reaction in kJ/mol.

$\Delta H=\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat released = 0.0238 kJ

n = number of moles NH₄NO₃ = $\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}=\frac{1.50g}{80g/mol}=0.01875mole$

$\Delta H=\frac{0.0238kJ}{0.01875mole}=1.27kJ/mole$

Therefore, the heat of the reaction is, 1.27 kJ/mole

8 0

3 years ago

ook at sample problem 18.12 in the 8th ed Silberberg book. Write a balanced chemical equation (salt hydrolysis). So acetate ion

vfiekz [6]

Answer:

Here's what I get

Explanation:

1. Write the chemical equation

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵

Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

$K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}$

3. Set up an ICE table

A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹: 0.35 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.35-x x x

4. Solve for x

$\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}$

Check for negligibility,

$\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}$

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

pH = 9.12

Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

3 0

4 years ago