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2 years ago

Which equation shows the proper setup to find the volume of a 0.250 M CaCl2 solution that contains 39.5 g of CaCl2?

Chemistry

2 answers:

MAVERICK [17]2 years ago

5 0

Answer: 1.42 L

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{\text{given mass of compound}}{\text{molar mass of compound}\times \text{volume of solution is L}}$

Given mass of $CaCl_2$ = 39.5 g

Molar mass of $CaCl_2$ = 111 g/mol

$0.250M=\frac{39.5g}{111}\times \text{volume of solution is L}}$

Volume of solution in L=1.42L

Jlenok [28]2 years ago

4 0

Molarity = number of mole of substance(n) / volume of solution (V).

n(CaCl2) = mass (CaCl2)/M(CaCl2)

M(CaCl2) = 40+2*35.5 = 111 g/mol

n(CaCl2) =39.5 g CaCl2*1 mol/111g

0.250 M = 39.5 g CaCl2*1 mol/111g*volume of solution (V).

volume of solution (V) = 39.5 g CaCl2*1 mol/(0.250 M*111g) = 1.42 L

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Gases emitted during volcanic activity often contain high concentrations of hydrogen sulfide and sulfur dioxide. these gases may

kompoz [17]

There can be three possible answers to this question: the amount of moles of SO₂ gas needed to react with 6.41 mol H₂S, and the amount of S and H₂O gas produced.

Amount of SO₂:
6.41 mol H₂S (1 mol SO₂/2 mol 2 mol H₂S) = 3.205 moles SO₂ gas

Amount of S:
6.41 mol H₂S (3 mol S/2 mol 2 mol H₂S) = 9.615 moles S solid

Amount of H₂O:
6.41 mol H₂S (2 mol H₂O/2 mol 2 mol H₂S) = 6.41 moles H₂O gas

7 0

2 years ago

Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?

balu736 [363]

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

8 0

2 years ago

In the formula X2O5, the symbol X could represent an element in Group

topjm [15]

Answer: (3) 15

Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.

6 0

2 years ago

An intravenous solution of mannitol is used as a diuretic to increase the loss of sodium and chloride by a patient. If a patient

Whitepunk [10]

Answer:

Explanation:

25 % ( m / v ) mannitol solution means 25 gram in 100 cc of water

25 gram in 100 mL of water

grams of mannitol in 30 mL = 25 x 30 / 100

= 7.5 grams .

6 0

2 years ago

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc = 3.74*10⁻³

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = 3.74*10⁻³

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago