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3 years ago

the mass concentration the mass concentration for a solution containing 45 g of calcium carbonate is 100 cm3

Chemistry

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer: The mass concentration for a solution containing 45 g of calcium carbonate for 100 $cm^{3}$ is 0.045 M.

Explanation:

Given: Mass of calcium carbonate = 45 g

Volume = $100 cm^{3}$

Convert $cm^{3}$ into L as follows.

$1 cm^{3} = 0.001 L\\100 cm^{3} = 100 cm^{3} \times \frac{0.001 L}{1 cm^{3}}\\= 0.1 L$

Moles of calcium carbonate (molar mass = 100 g/mol) is as follows.

$No. of moles = \frac{45 g}{100 g/mol}\\= 0.45 mol$

Molarity is the number of moles of solute present in a liter of solution.

Hence, molarity (or mass concentration) of given solution is calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.45 mol}{0.1 L}\\= 0.045 M$

Thus, we can conclude that the mass concentration for a solution containing 45 g of calcium carbonate for 100 $cm^{3}$ is 0.045 M.

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Explanation:

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2 years ago

The hydrogen gas produced is collected over water at 25.0 oC. The volume of the gas is 7.80 L, and the total pressure in the gas

Eva8 [605]

Answer:

20.4 grams Zn

Explanation:

To find the mass, you first need to find the moles. This can be found using the Ideal Gas Law equation:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the values into the equation, you need to convert Celsius to Kelvin.

P = 0.980 atm R = 0.08206 atm*L/mol*K

V = 7.80 L T = 25.0 °C + 273.15 = 298.15 K

n = ? moles

PV = nRT

(0.980 atm)(7.80 L) = n(0.08206 atm*L/mol*K)(298.15 K)

7.644 = n(24.466)

0.312 moles = n

Now that you have the number of moles, you can convert it to grams using the atomic mass of zinc. The final answer should have 3 sig figs to match the sig figs in the given values.

Atomic Mass (Zn): 65.380 g/mol

0.312 moles Zn 65.380 grams
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5 0

1 year ago

I need help with this, please :00000

mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if 1 mole of N₂ produce 2 moles of NH₃

then 3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

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limiting reactant

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3 years ago

Use the kinetic theory to explain why a windy day is a good drying day?

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2 years ago

Read 2 more answers

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.

4 0

3 years ago

the mass concentration the mass concentration for a solution containing 45 g of calcium carbonate is 100 cm3 ​

the mass concentration the mass concentration for a solution containing 45 g of calcium carbonate is 100 cm3