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3 years ago

Predict the density of acetylene gas (C2H2) at 0.910 atm and 20oC.

Chemistry

1 answer:

zhuklara [117]3 years ago

6 0

Answer:

d = 0.98 g/L

Explanation:

Given data:

Density of acetylene = ?

Pressure = 0.910 atm

Temperature = 20°C (20+273 = 293 K)

Solution:

Formula:

PM = dRT

R = general gas constant = 0.0821 atm.L/mol.K

M = molecular mass = 26.04 g/mol

0.910 atm × 26.04 g/mol = d × 0.0821 atm.L/mol.K×293 K

23.7 atm.g/mol = d × 24.1 atm.L/mol

d = 23.7 atm.g/mol / 24.1 atm.L/mol

d = 0.98 g/L

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If you created a food chain using the following items. which item would follow the corn?

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3 years ago

Read 2 more answers

Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

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Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

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3 years ago

How many particles are in 23 g of H 2 O?

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1 mole of any substance contains 6.022 × 1023 particles.

⚛ 6.022 × 1023 is known as the Avogadro Number or Avogadro Constant and is given the symbol NA

N = n × NA

· N = number of particles in the substance

· n = amount of substance in moles (mol)

· NA = Avogardro Number = 6.022 × 10^23 particles mol-1

For H2O we have:

2 H at 1.0 each = 2.0 amu
1 O at 16.0 each = 16.0 amu
Total for H2O = 18.0 amu, or grams/mole

It takes 18 grams of H2O to obtain 1 mole, or 6.02 x 1023 molecules of water. Think about that before we answer the question. We have 25.0 grams of water, so we have more than one mole of water molecules. To find the exact number, divide the available mass (25.0g) by the molar mass (18.0g/mole). Watch how the units work out. The grams cancel and moles moves to the top, leaving moles of water. [g/(g/mole) = moles].

Here we have 25.0 g/(18.0g/mole) = 1.39 moles water (3 sig figs).

Multiply 1.39 moles times the definition of a mole to arrive at the actual number of water molecules:

1.39 (moles water) * 6.02 x 1023 molecules water/(mole water) = 8.36 x 1023 molecules water.

That's slightly above Avogadro's number, which is what we expected. Keeping the units in the calculations is annoying, I know, but it helps guide the operations and if you wind up with the unit desired, there is a good chance you've done the problem correctly.

N = n × (6.022 × 10^23)

1 grams H2O is equal to 0.055508435061792 mol.

Then 23 g of H2O is 1.2767 mol

To calculate the number of particles, N, in a substance:

N = n × NA

N = 1.2767 × (6.022 × 10^23)

N= 176.26

N=

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ss7ja [257]

Answer:

An element with 7 valence electrons will most likely be a halogen and gain an electron

Explanation:

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