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allochka39001 [22]

3 years ago

11

Select the correct answer. Thomas has 235 grams of K2S in the chemistry lab. How many atoms of potassium (K) are in 235 grams of

the compound? A. 2. 13 × 1023 B. 4. 26 × 1023 C. 2. 57 × 1024 D. 3. 37 × 1024 E. 7. 08 × 1024.

1 answer:

Readme [11.4K]3 years ago

6 0

Atoms are the smallest division of the element. The potassium atoms present in the 235 gm of the compound is $2.57 \times 10^{24}$ atoms.

<h3>What is the number of atoms?</h3>

Given,

Mass (m) of Potassium sulfide $(\rm K_{2}S)$ = 235 gm
Molar mass (M) of Potassium sulfide = 110.26 g/mol

Calculate the number of moles as:

$\begin{aligned}\rm Moles (n) &= \dfrac{\rm mass}{\rm molar\;\rm mass}\\\\&= \dfrac{235}{110.26}\\\\&=2.13\;\rm mol\end{aligned}$

In the given compound $\rm K_{2}S$ , there are two atoms of potassium and one atom of sulphur.

If, 1 mole = $2 \times 6.022 \times 10^{23}$ atoms of potassium

Then, 2.13 moles = X atoms

Solving for X:

$2.13 \times 2 \times 6.022 \times 10^{23} = 2.57 \times 10^{24}$

Therefore, option c. $2.57 \times 10^{24}$ is correct.

Learn more about atoms here:

brainly.com/question/11411852

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snow_tiger [21]

Answer:

inertia

Explanation:

5 0

3 years ago

Describe the chemical reaction of: <br> NH3(g)+O2(g) -> NO(g)+ H2O

Pie

Answer:

Oxidation of ammonia

Explanation:

Chemical equation:

NH₃ + O₂ → NO + H₂O

Balanced chemical equation:

4NH₃ + 5O₂ → 4NO + 6H₂O

The given reaction shows the oxidation ammonia. When oxygen react with ammonia it form nitrogen monoxide and water vapors. Nitrogen monoxide decomposes in air.

It can be seen that the given reaction also follow the law of conservation of mass. There are four nitrogen, ten oxygen and twelve hydrogen atoms on both side of equation.

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

7 0

3 years ago

A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of $0.80^oC$ and a freezing point depression constant $K_f=7.82^oC.kg/mol$ . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is $-17.6^oC$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $0.80^oC$

Freezing point of solution = $?^oC$

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

$K_f$ = freezing point depression constant = $7.82^oC/m$

$m_{solute}$ = Given mass of solute (iron (III) chloride) = 81.1 g

$M_{solute}$ = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

$w_{solvent}$ = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

$0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC$

Hence, the freezing point of the solution is $-17.6^oC$

6 0

3 years ago

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

3 years ago

Enter your answer in the provided box. Consider the reaction: N2(g) + 3H2(g) → 2NH3(g) Suppose that a particular moment during t

Mashcka [7]

Answer:

0.0253 M/s

Explanation:

From the reaction

N₂ + 3H₂ → 2NH₃

The rate of reaction can be written as

Rate = - $\frac{d[N_2]}{dt}$ = - $\frac{1}{3} \frac{d[H_2]}{dt}$ = + $\frac{1}{2} \frac{d[NH_3]}{dt}$

From the above rate equation we can conclude that the rate of reaction of N₂ is equal to one third of the rate of reaction of H₂,

So,

Rate of reaction of molecular nitrogen = $\frac{1}{3} \times 0.0759$

Upon calculation, we get rate of reaction of molecular nitrogen = 0.0253 M/s

4 0

3 years ago