which of the following does not represent a molecular compound? A)CS2 B)Bal2 C)N2O4 D)PCI3

If .00327 g of a gas dissolves in 0.376 L of water at 867 torr, what quantity of this gas (in grams) will dissolve at 759 torr?

Vinil7 [7]

At 759 torr, 0.002 86 g of the gas will dissolve.

Henry’s law states:

$c = k_{H}p$ , where

where $K_{H}$ is a proportionality constant called the Henry's Law constant.

If we have the same solute in the same solvent at two different pressures $p_{1}$ and $p_{2}$ ,

$c_{1} = k_{H}p_{1}$ and $c_{2} = k_{H}p_{2}$

Dividing the two equations, k_H cancels and we get

c_1/c_2 = p_1/p_2

c_2 = c_1 × p_2/p_1

The volumes of solvent are the same, so we can use the masses of the solute instead of concentrations.

∴ c_2 = 0.003 27 g × (759 torr/867 torr) = 0.002 86 g

8 0

3 years ago

Which of the following is not true in regard to noble gases? A. These elements have full outermost shells. B. These elements are

Likurg_2 [28]

Answer:

D.These elements tend to gain or lose electrons easily is incorrect.

Explanation:

Noble gasses consist of atoms that have full rings of electrons. Electrons are what bond atoms together to create chemical compounds, and incomplete rings are needed for the process.

5 0

3 years ago

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Hi there, giving brainliest to best answer. no spammers

Juli2301 [7.4K]

Answer:

D. Earth's northern hemisphere is tilted toward the Sun

Explanation:

You can see in the picture above that the northern hemisphere is past the perpendicular to orbit line to the right.

4 0

2 years ago

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What is the density of chlorine gas (MM = 71.0glmol) at 1.50 atm and 25.0C

IrinaVladis [17]

Answer: D=4.35g/L

Explanation:

The formula for density is $D=\frac{M}{V}$ . M is mass in grams and V is volume in liters.

Since we are give pressure and temperature, we can use the ideal gas law to find moles/volume. FInding moles/volume would give us the base for density. All we would have to do is convert moles to grams.

Ideal Gas Law: PV=nRT

$\frac{n}{V} =\frac{P}{RT}$

$\frac{n}{V}=\frac{1.50 atm}{(0.08206Latm/Kmol)(25+274.15K)}$

$\frac{n}{V} =\frac{0.061309mol}{L}$

Now that we have moles, we can use molar mass of chlorine gas to find grams.

$0.061309mol*\frac{71.0g}{mol} =4.3529g$

With our grams, we can find our density.

$D=\frac{4.3529g}{L}$

We need correct significant figures so our density is:

$D=\frac{4.35g}{L}$

5 0

3 years ago

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

Step 1: Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

Step 2: Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

Step 3: Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

Step 7: Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

Step 8: Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0

3 years ago