If the density of carbon tetrachloride is 0.893 g/mL, and a sample has a volume of 9.29 mL, what is the mass?

620 mL of nitrogen at standard pressure is compressed into a 480 mL container. What is the new pressure in kPa?

bulgar [2K]

Answer:

c. 131 kPa

Explanation:

Hello!

In this case, since the relationship between volume and pressure is inversely proportional, based on the Boyle's law:

$P_1V_1=P_2V_2$

Considering that the standard pressure is 101.325 kPa, we can compute the final pressure as shown below:

$P_2=\frac{P_1V_1}{V_2}=\frac{620mL*101.325kPa}{480mL}\\\\P_2=131kPa$

Therefore, the answer is c. 131 kPa .

Best regards!!

7 0

2 years ago

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

vazorg [7]

C + O2= CO2
$n = \frac{m}{mw}$
$n = \frac{5.4}{12} \\ n = 0.45 \: mol \: of \: carbon$
$n = \frac{13.6}{12 + 16 \times 2} \\ n = \frac{13.6}{44} \\ n = 0.31 \: mol \: of \: carbon \: dioxide$
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %

4 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

What are the answers please help

LenaWriter [7]

Thank you for the free 15 points .

4 0

2 years ago

The following equilibrium is formed when copper and bromide ions are placed in a solution:

JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

$heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)$

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0

3 years ago