HELP!!! name two properties of metals

a sample of sulfur dioxide occupies a volume of 652 mL at 40.0 C and 0.75 atm. What volume will the sulfur dioxide occupy at STP

sesenic [268]

The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the following formula:

PV = nRT

P = pressure
V = volume
n = number of moles
R = gas law constant
T = temperature

0.75 × 0.652 = n × 0.0821 × 313

0.489 = 25.69n

n = 0.489/25.69

n = 0.019moles

Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.

Learn more about volume at: brainly.com/question/1578538

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5 0

1 year ago

Which of the following types of bonds will have the highest electrical conductivity?

masha68 [24]

I think your answer is A not sure tho

8 0

1 year ago

Read 2 more answers

Identify the mixture of powdered charcoal and powdered sugar and suggest a technique for separating their components

mihalych1998 [28]

A- Identify the mixture:
The mixture of powdered charcoal and powdered sugar is considered as a homogeneous mixture. This means that you cannot identify the components with naked eye as they are uniformly distributed in the mixture.

B- Separate components:
You ca separate the charcoal powder from the sugar powder using the following steps:
1- add water. Sugar will dissolve in water while charcoal won't.
2- filter the solution where the powdered charcoal will remain on the filter paper and the solution of powder will pass through.
3- boil the sugar solution (above 100 degrees celcius). The water will evaporate and the sugar will precipitate.

5 0

2 years ago

Given the standard heats of reaction

ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole

6 0

2 years ago

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.

ruslelena [56]

Answer:

a) 0.168 moles O2

b) 9.50 grams O2

c) 0.01662 kg NO

d)88.9 %

Explanation:

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

a. How many moles of ammonia will react with 6.73g of oxygen?

Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen = 6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.210 moles O2 we need 4/5 *0.210 = 0.168 moles O2

b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

Mass O2 = 0.297 moles O2 * 32.00 g/mol = 9.50 grams O2

c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

Calculate moles of ammonia = 9.43105 grams / 17.03 g/mol =0.5538 moles

Calculate moles of NO:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.5538 moles of NH3 we'll have 0.5538 moles NO

Calculate mass of NO

Mass NO = 0.5538 moles * 30.01 g/mol = 16.62 grams = 0.01662 kg NO

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

Calculate moles of NH3 = 2.51 grams / 17.03 g/mol = 0.147 moles

Calculate moles of O2 = 3.76 grams / 32 g/mol = 0.118 moles

Determine the limiting reactant

O2 is the limiting reactant, it will completely be consumed (0.118 moles)

NH3 is in excess. There will react 4/5 * 0.118 = 0.0944 moles

There will remain 0.147 - 0.0944 = 0.0526 moles

Calculate moles H2O: For 0.118 moles O2 we'll have 6/5 * 0.118 = 0.1416 moles H2O

Calculate mass H2O = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

Calculate % yield = (2.27/2.552)*100 % = 88.9 %

4 0

2 years ago