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3 years ago

What is the volume of a gas a t320 K, if the gas occupies 50.0 mL at a temperature of 273 K?

Chemistry

1 answer:

shtirl [24]3 years ago

6 0

Answer:

58.61mL

Explanation:

V1 (initial volume) =?

T1 (initial temperature) = 320 K

V2 (final volume) = 50mL

T2 (final temperature) = 273 K

Using the Charles' law equation V1/T1 = V2/T2

The initial volume of the gas can obtained as follow:

V1/320 = 50/273

Cross multiply to express in linear form

V1 x 273 = 320 x 50

Divide both side by 273

V1 = (320 x 50)/273

V1 = 58.61mL

Therefore, the initial volume of the gas is 58.61mL

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3 years ago

Mg + 2HCl → MgCl2 + H2

deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

2HCl(aq)

→

MgCl

(

)

+ H

(

)

Determine Moles of Magnesium

Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

g Mg

mol Mg

24.3050

g Mg

0.200 mol Mg

Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

100 mL

and then to

0.1 L

1 dm

1 L

Convert

2.00 mol/dm

2.00 mol/L

Multiply

0.1

times

2.00 mol/L

100

1000

0.1 L HCl

2.00 mol/dm

2.00 mol/L

0.1

2.00

mol

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

2.01588 g/mol

0.200

mol Mg

mol H

mol Mg

2.01588

g H

mol H

0.403 g H

0.200

mol HCl

mol H

mol HCl

2.01588

g H

mol H

0.202 g H

The limiting reactant is

HCl

, which will produce

0.202 g H

under the stated conditions.

pls mark as brainliest ans

7 0

2 years ago

Read 2 more answers

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago

A slice of Swiss cheese contains 48 mg of sodium

Nookie1986 [14]

Answer: grams=0.048g, ounces=0.0017oz, 0.00011lb

Explanation:

Stoichiometry

48 mg x 1 g

÷ 1000 mg = 0.048 g

48 mg x 1 g x 16 oz

÷ 1000 mg ÷ 453.6 g = 0.0017 oz

48 mg x 1 g x 1 lb

÷ 1000 mg ÷ 453.6 g = 0.00011 lb

6 0

3 years ago

In the name, iron(III) oxide, the (III) represents

miss Akunina [59]

Answer:

Explanation:

the number o iron atoms

7 0

3 years ago