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3 years ago

Find the volume of 20.0 g of benzene. It's density is 0.88g/mL

1 answer:

7 0

$d=0,88\frac{g}{mL}\\
m=20g\\\\
d=\frac{m}{V} \ \ \ \Rightarrow \ \ \ V=\frac{m}{d}\\\\\
V=\frac{20g}{0,88\frac{g}{mL}}\approx22,73mL$

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Which substance is a heterogeneous mixture

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3 years ago

Read 2 more answers

9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

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1 year ago

7th grade science help me plsssss

tigry1 [53]

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3 years ago

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

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3 years ago

What is the main difference between active and passive transport?

masha68 [24]

Answer:

active transport requires energy and passive transport does not require energy

Explanation:

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2 years ago