Determine the rate of reaction based on the data below:

Density is the ratio of a sample's mass to its volume. A bar of lead has a mass of 115.2 g. When it is submerged in 25.0 mL of w

Margarita [4]

Answer:

$10.97 \ g/cm^3$

Explanation:

Given :

Mass of a bar of lead = 115.2 g

Initial water level $\text{in the graduated cylinder}$ = 25 mL

Final water level $\text{in the graduated cylinder}$ = 35.5 mL

Difference in the water level = 35.5 - 25

= 10.5 mL

= $10.5 \ cm^3$

We know that when a body is submerged in water, it displaces its own volume of water.

Therefore, the volume of the lead bar = volume of the water displaced = 10.5 mL = $10.5 \ cm^3$

We know that mathematically, density is the ratio of mass of body to its volume.

Density of the lead bar is given by :

$\rho =\frac{\text{mass}}{\text{volume}}$

$\rho =\frac{\text{115.2 g}}{\text{10.5 cm}^3}$

= $10.97 \ g/cm^3$

3 0

3 years ago

A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou

Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0

3 years ago

How many particles would be found in a 89.2 g sample of Na?

julsineya [31]

One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert 89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles

5 0

3 years ago

Polyatomic ion formula for aluminum hydroxide

KonstantinChe [14]

Answer:

Al(OH)3 ? i hope this is what you mean.

7 0

3 years ago

Read 2 more answers

15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what

s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.

M_r: 52.00 80.91 291.71

2Cr + 6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g: 15.00 15.00

Step 2. Calculate the moles of each reactant

Moles of Cr = 15.00 × 1/52.00

Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×

Step 3. Identify the limiting reactant

Calculate the moles of CrCl₃ we can obtain from each reactant.

From Cr:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

From HBr:

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

Step 4. Calculate the theoretical yields of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

Theoretical yield of H₂ = 0.1854 × 3/6

Theoretical yield of H₂ = 0.092 70 mol H₂

Step 5. Calculate the volume of H₂ at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

Step 6. Calculate the percent yield

% Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

% yield = 15.35/18.03 × 100

% yield = 85.15 %

8 0

3 years ago