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2 years ago

Need answer now! 100 points

Chemistry

2 answers:

ratelena [41]2 years ago

5 0

Answer: D

divide the moles of the substances given in the question by the co-efficient in the equation. Smallest quotient is the limiting reactant

s344n2d4d5 [400]2 years ago

3 0

Answer:

D) HBr

Explanation:

Given data:

Number of moles of Al = 8 mol

Number of moles of HBr = 8 mol

Limiting reactant = ?

Solution:

2Al + 6HBr → 2AlBr₃ + 3H₂

now we will compare the moles of reactants with products.

Al : AlBr₃

2 : 2

8 : 8

Al : H₂

2 : 3

8 : 3/2×8 = 12

HBr : AlBr₃

6 : 2

8 : 2/6×8 = 2.67

HBr : H₂

6 : 3

8 : 3/6×8 = 4

HBr produced less number of moles of product thus it will act as limiting reactant.

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A 0.08541 g sample of gas occupies 10.0-ml at 288.5 k and 1.10 atm. upon further analysis, the compound is found to be 13.068% c

topjm [15]

<span>C2Br2 First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is PV = nRT where P = pressure (1.10 atm = 111458 Pa) V = volume (10.0 ml = 0.0000100 m^3) n = number of moles R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) ) T = Absolute temperature Solving for n, we get PV/(RT) = n Now substituting our known values into the formula. (111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol)) = (1.11458/2398.721652) mol = 0.000464656 mol Now let's calculate the empirical formula for this compound. Atomic weight carbon = 12.0107 Atomic weight bromine = 79.904 Relative moles carbon = 13.068 / 12.0107 = 1.08802984 Relative moles bromine = 86.932 / 79.904 = 1.087955547 So the relative number of atoms of the two elements is 1.08802984 : 1.087955547 After dividing all numbers by the smallest, the ratio becomes 1.000068287 : 1 Which is close enough to 1:1 for me to consider the empirical formula to be CBr Now calculate the molar mass of CBr 12.0107 + 79.904 = 91.9147 Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So 91.9147 g/mol * 0.000464656 mol = 0.042708701 g 0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is. 0.08541 / 0.0427087 = 1.99982673 1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>

3 0

3 years ago

Read 2 more answers

when carbon is burned in the air, it reacts with oxygen to form carbon dioxide. when 22.8 g of carbon were burned in the presenc

OverLord2011 [107]

Answer:

So there is83.6g CO2 produced

Explanation:

Burning carbon with air has the following equation

C + O2 → CO2

For 1 mol Carbon, we have 1 mol O2 and 1 mol CO2

Step 2: Calculating moles

mole C = 22.8g / 12g/mole

Mole C = 1.9 mole

1.9 mole C will completely react

Since for each mole C there is 1 mole O2 and 1 mole CO2

This means there will also react 1.9 mole of 02, to be formed 1.9 mole of CO2

mole CO2 = mass CO2 / Molar mass CO2

mass CO2 = 1.9 mole CO2 * 44g/mole =<u>83.6g CO2</u>

In this reaction 18.2 g of O2 remained unreacted

we can control this: 79g - 18.2 g = 60.8g

1.9 mole * 32g/mol = 60.8g

So there is83.6g CO2 produced

4 0

3 years ago

What is the volume of a balloon of gas at 842 mm Hg and -23° C, if its volume is 915 mL at a pressure of 1,170 mm Hg and a tempe

garik1379 [7]

Answer:
V₂ = 1070 mL or 1.07 L

Solution:

Data Given;
P₁ = 1170 mmHg

V₁ = 915 mL

T₁ = 24 °C + 273 K = 297 K

P₂ = 842 mmHg

V₂ = ?

T₂ = - 23 °C + 273 K = 250 K

According to Ideal gas equation,

P₁ V₁ / T₁ = P₂ V₂ / T₂

Solving for V₂,

V₂ = P₁ V₁ T₂ / P₂ T₁

Putting Values,

V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)

V₂ = 1070 mL or 1.07 L

5 0

3 years ago

Based on the kinetic molecular theory, which of the following statements is correct about a sample of gas at a constant temperat

Scilla [17]

Answer:

C) It has a constant average kinetic energy

Explanation:

The average kinetic energy of the particles in a gas is directly proportional to the temperature of the gas, according to the equation.

k is the Boltzmann's constant

T is the absolute temperature of the gas

Therefore, temperature of a gas is a measure of the average kinetic energy of the particles.

In this problem, we are told that the gas is at constant temperature (and volume): therefore, according to the previous equation, this means that the average kinetic energy is also constant.

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3 years ago

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enyata [817]

Answer:

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6 0

3 years ago