How many grams of CO2 could be formed with 2.09 x 1023 atoms of O?

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

3 years ago

Read 2 more answers

A chemist designs a galvanic cell that uses these two half-reactions:

Virty [35]

Answer:

MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)

Explanation:

a)

MnO4⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn(aq)²⁺ + 4H2O (l)

b)

5Fe³⁺ (aq) +5e⁻ → 5Fe²⁺(aq)

c)

MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)

8 0

2 years ago

What is the total number of oxygen atoms represented in the formula na2c03.10h2o?

Wewaii [24]

As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence

Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13

number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6

Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus

number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23

number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23

= 1.566 X 10^24

Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
4.7

7 0

3 years ago

This is what your screen should look like. start with Hydrogen -1 and drag 2 neutrons over the to the nucleus. What happens to t

Alekssandra [29.7K]

Answer:

A tritium is produced.

Explanation:

Combining two additional neutrons to the nucleus of the hydrogen atom makes it a tritium, Hydrogen-3.

neutron is designated ¹₀n; this shows a mass number of 1 and no atomic number

Hydrogen-1 is designated as ₁¹H; a mass number of 1 and atomic number of 1. This particle is actually more like a proton.

Combining both:

₁¹H + 2¹₀n → ³₁H

This is a nuclear reaction and in balancing such reaction equation, mass numbers and atomic numbers must be conserved.

4 0

3 years ago

When 16.0 g of an unknown compound (a nonelectrolyte) are dis solved in exactly 800. g of water, the solution has a freezing poi

dexar [7]

Answer:

A. 266g/mol

Explanation:

A colligative property of matter is freezing point depression. The formula is:

ΔT = i×Kf×m (1)

Where:

ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g

Replacing in (1):

0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg

0,0614 = mol of solute.

As molar mass is defined as grams per mole of substance and the compound weights 16,0g:

16,0g / 0,0614 mol = 261 g/mol ≈ A. 266g/mol

I hope it helps!

3 0

2 years ago