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3 years ago

How many grams of KCl is needed to make .75 L of a 1 M solution of KCl?

Chemistry

1 answer:

Zielflug [23.3K]3 years ago

3 0

Answer:

The answer to your question is 25.9 g of KCl

Explanation:

Data

Grams of KCl = ?

Volume = 0.75 l

Molarity = 1 M

Formula

$Molarity = \frac{number of moles}{volume}$

Solve for number of moles

$Number of moles = Molarity x volume$

Substitution

Number of moles = 1 x 0.75

Simplification

Number of moles = 0.75 moles

Molecular mass KCl = 39 + 35.5 = 34.5

Use proportions to find the grams of KCl

34.5 g of KCl ---------------- 1 mol

x ---------------- 0.75 moles

x = (0.75 x 34.5) / 1

x = 25.9 g of KCl

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2Na3N3 -> 2Na + 3N2<br><br> if 0.58 mol if NaN3 reacts, what mass of nitrogen would result

Ronch [10]

Answer:

Mass = 24.36 g of N₂

Explanation:

The balance chemical equation for the decomposition of NaNO₃ is as follow;

2 NaN₃ → 2 Na + 3 N₂

Step 1: Find moles of N₂ as;

According to equation,

2 moles of NaNO₃ produces = 3 moles of N₂

So,

0.58 moles of NaNO₃ will produce = X moles of N₂

Solving for X,

X = 3 mol × 0.58 mol / 2 mol

X = 0.87 mol of N₂

Step 2: Calculate mass of N₂ as,

Mass = Moles × M.Mass

Mass = 0.87 mol × 28.01 g/mol

Mass = 24.36 g of N₂

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3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

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3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

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3 years ago

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