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3 years ago

a student needs to use 0.575 moles of sodium chloride in an experiment. How many grams should he mass out?

Chemistry

1 answer:

Bad White [126]3 years ago

5 0

The molecular weight of NaCl is 58.44 g/mol.

0.575 mol * 58.44 g/mol = 33.6 grams of NaCl

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A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai

marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

C: 12.011;
O: 15.999.

Calculate the molar mass of carbon dioxide $\rm CO_2$ :

$M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

Find the number of moles of molecules in that $41.1\;\rm g$ sample of $\rm CO_2$ :

$n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol$ .

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

$P \cdot V = n \cdot R \cdot T$ ,

where

$P$ is the pressure inside the container.
$V$ is the volume of the container.
$n$ is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
$R$ is the ideal gas constant.
$T$ is the absolute temperature of the gas.

Rearrange the equation to find an expression for $P$ , the pressure inside the container.

$\displaystyle P = \frac{n \cdot R \cdot T}{V}$ .

Look up the ideal gas constant in the appropriate units.

$R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}$ .

Evaluate the expression for $P$ :

$\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}$ .

Apply dimensional analysis to verify the unit of pressure.

4 0

3 years ago

Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 gr

Lunna [17]

Answer: Bromine is the limiting reactant

Explanation:

First of all let's generate a balanced equation for the reaction

2Li + Br2 —> 2LiBr

Molar Mass of Li = 7g/mol

Molar Mass of Br2 = 2x80 = 160g/mol

From the question given, were told that 25g of Li and 25g Br2 were present at the take-off of the reaction. Converting these Masses to mole, we have:

Number of mole of Li = 25/7 = 3.6moles

Number of mole of Br2 = 25/160 = 0.156mol.

To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

For the experimental yield:

Li : Br2 = 3.6/ 0.156 = 23 : 1

For the theoretical yield:

Li : Br = 2 : 1

From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.

4 0

3 years ago

PLEASE HELP MEEEE ASAP 50 PTS

belka [17]

Answer: The answerer is CH4 + 2O2 → CO2 + 2H2O

Explanation: }

7 0

2 years ago

Read 2 more answers

Chemical bonding is the combining of atoms to form ______________?

svet-max [94.6K]

A i believe its atoms