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Elanso [62]
4 years ago
12

A student measures a pressure increase of 0.25 atm while collecting the H2 gas products from the reaction of Mg with HCl. The th

eoretical pressure change is 0.29 atm. What is the % yield
Chemistry
1 answer:
natulia [17]4 years ago
7 0

Answer: Thus the % yield is 86.2%

Explanation:

Percentage yield is the ratio of actual change to the theoretical change in terms of percentage.

The given reaction is :

Mg+2HCl\rightarrow MgCl_2+H_2

To calculate the percent yield :

\%\text{ yield of reaction}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Given : Actual pressure increase = 0.25 atm

Theoretical pressure change = 0.29 atm.

\%\text{ yield of reaction}=\frac{0.25}{0.29}\times 100=86.2\%

Thus the % yield is 86.2%

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Rudik [331]

Answer:

The photoelectric effect

Explanation:  hope it helps

5 0
3 years ago
The rate constant for this second‑order reaction is
o-na [289]

Answer:

daddadaddddddddddddddddddddddddddddda

Explanation:

dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

3 0
3 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0
3 years ago
If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
frutty [35]

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

3 0
4 years ago
Mass of metal=0.0291. Volume of gas collected over water=25.67mL. Temperature=24.1C. Atmospheric pressure=754.6mmHg. Note: Metal
NISA [10]

Answers:

1) 732.1 mmHg; 2) 0.001 014 mol; 3) 0.001 014 mol; 4) 28.7 g/mol; 5) 187 ppt.

Explanation:

1) <em>Partial pressure of hydrogen</em>

You are collecting the gas over water, so

p_{\text{atm}} = p_{\text{H}_{2}} + p_{\text{H}_{2}\text{O}}

p_{\text{H}_{2}} = p_{\text{atm}} - p_{\text{H}_{2}\text{O}}

p_{\text{atm}} = \text{754.6 mmHg}

At 24.1 °C, p_{\text{H}_{2}\text{O}} = \text{22.5 mmHg}

p_{\text{H}_{2}} = \text{754.6 mmHg} - \text{22.5 mmHg} = \textbf{732.1 mmHg}

===============

2) Moles of H₂

We can use the Ideal Gas Law.

<em>pV = nRT</em>                Divide both sides by <em>RT</em> and switch

<em>n</em> = (<em>pV</em>)/(<em>RT</em>)

p = 732.1 mmHg                                                 Convert to atmospheres

p = 732.1/760                                                      Do the division

p = 0.9633 atm

V = 25.67 mL                                                        Convert to litres

V = 0.025 67 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 24.1 °C                                                              Convert to kelvins

T = (24.1 + 273.15 ) K = 297.25 K                          Insert the values

n = (0.9633 × 0.025 67)/(0.082 06 × 297.25)     Do the multiplications

n = 0.02473/24.39                                                 Do the division

n = 0.001 014 mol

===============

3)<em> Moles of metal </em>

The partial chemical equation is

M + … ⟶ H₂ + …

The molar ratio of M:H₂ is 1 mol M:1 mol H₂.

Moles of M = 0.001 014 × 1/1                          Do the operations

Moles of M = 0.001 014 mol M

===============

4) Atomic mass of M

Atomic mass = mass of M/moles of M     Insert the values

Atomic mass = 0.0291/0.001 014             Do the division

Atomic mass = 28.7 g/mol

===============

5) <em>Relative deviation in ppt </em>

Your metal must be in Group 2 because of the 1:1 molar ratio of M:H₂.

The metal with the closest atomic mass is Mg (24.305 g/mol).

Relative deviation in ppt = |Experimental value – Theoretical value|/Theoretical value × 1000

Relative deviation = |28.7 – 23.405|/23.405 × 1000     Do the subtraction

Relative deviation = |4.39|/23.405 × 1000                      Do the operations

Relative deviation = 187 ppt

3 0
3 years ago
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