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2 years ago

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In the previous problem, which are the limiting reactants for each reaction? Based on this, calculate the maximum yield of oxyge

n, carbon dioxide, and sulfur dioxide when 280 g of black powder is burned.

1 answer:

juin [17]2 years ago

4 0

please show the previous problem

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What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water?

dusya [7]

Answer:

The concentration of KBr is $C = 0.07899 \ mol L^{-1}$

Explanation:

From the question we are told that

The mass of KBr is $m_{KBr} = 2.35 \ g$

The molar mass of KBr is $M_{KBr} = 119 g/mol$

Volume of water is $V = 250 \ mL = 250 *10^{-3} = 0.250 \ L$

This implies that the volume of the solution is $V = 250 mL$

The number of moles of KBr is

$n = \frac{m_{KBr}}{M_{KBr}}$

Substituting values

$n = \frac{2.35}{119}$

$n = 0.01975 \ mol$

The concentration of KBr is mathematically represented as

$C = \frac{0.01975}{0.250}$

$C = 0.07899 \ mol L^{-1}$

3 0

2 years ago

A single 250 gram

Gala2k [10]

2.2 x 10^-2

0.055 / 250 = 0.00022 - This would be 2.2 x 10^-4, but the question is asking for percent, not proportion, so multiply by 100% to get the percentage.

0.00022 * 100% = 0.022% = 2.2 * 10^-2

6 0

2 years ago

How many molecules are in 2.50 moles of carbon dioxide?

podryga [215]

Answer:

1.5055×10²⁴ molecules

Explanation:

From the question given above, the following data were obtained:

Number of mole CO₂ = 2.5 moles

Number of molecules CO₂ =?

The number of molecules present in 2.5 moles CO₂ can be obtained as:

From Avogadro's hypothesis,

1 mole of CO₂ = 6.022×10²³ molecules

Therefore,

2.5 mole of CO₂ = 2.5 × 6.022×10²³

2.5 mole of CO₂ = 1.5055×10²⁴ molecules

Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂

7 0

2 years ago

How do I calculate the sum of 6.078 g and 0.3329 g ?

Vesnalui [34]

The answer is 6.4109 grams

8 0

2 years ago

A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will b

Angelina_Jolie [31]

Answer:

50.4 % of the water will be vapor

Explanation:

<u>Step 1:</u> Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

<u>Step 2:</u> Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V)/(R*T)

n = (0.2467 * 5.3) /(0.08206* 338)

n = 0.047 moles

<u>Step 3:</u> Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles *18.02 g/mol

Mass of water = 0.84694 grams

<u>Step 4:</u> Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) *100%

% = 50.4%

50.4 % of the water will be vapor

5 0

2 years ago