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kykrilka [37]
3 years ago
7

How many moles of Na3PO4 would be required to react with 1.0 mol of AgNO3?

Chemistry
2 answers:
Solnce55 [7]3 years ago
6 0
1/3 mol or .3 repeating, I believe.
Tema [17]3 years ago
5 0

<u>Answer:</u> The number of moles of sodium phosphate required is 0.33 moles

<u>Explanation:</u>

We are given:

Moles of silver nitrate = 1 mole

The chemical equation for the reaction of sodium phosphate and silver nitrate follows:

Na_3PO_4+3AgNO_3\rightarrow 3NaNO_3+Ag_3PO_4

By Stoichiometry of the reaction:

3 moles of silver nitrate reacts with 1 mole of sodium phosphate.

So, 1.0 moles of silver nitrate will react with = \frac{1}{3}\times 1.0=0.33mol of sodium phosphate.

Hence, the number of moles of sodium phosphate required is 0.33 moles

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How much does the Earth weigh?​
ELEN [110]

5.972 × 10^24 kg

hope it helps

6 0
3 years ago
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Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol
Sonbull [250]

<u>Answer:</u> The value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

                N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>            1.30       1.65

<u>At eqllm:</u>       1.30-x    1.65-3x             2x

Evaluating the value of 'x'

\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol

The expression of K_c for above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

Equilibrium moles of nitrogen gas = (1.30-x)=(1.30-0.05)=1.25mol

Equilibrium moles of hydrogen gas = (1.65-x)=(1.65-0.05)=1.60mol

Putting values in above expression, we get:

K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}

Calculating the K_c' for the given chemical equation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2

Hence, the value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

8 0
2 years ago
Read 2 more answers
Which is NOT true of ionic bonding?
WARRIOR [948]

Answer:

D is wrong

Explanation:

it doesn't involves any kind of sharing of electrons

6 0
2 years ago
Information related to a titration experiment is given in the balanced equation and table below.
Scorpion4ik [409]

Answer:

0.24M

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the equation above, we obtained the following information:

nA (mole of acid) = 1

nB (mole of base) = 2

Data obtained from the question include:

Va (volume of the acid) = 12mL

Ca (concentration of the acid) =?

Vb (volume of the base) = 36mL

Cb (concentration of the base) = 0.16 M

The Ca (concentration of the acid) can be obtained as follow:

CaVa/CbVb = nA/nB

Ca x 12 / 0.16 x 36 = 1 /2

Cross multiply to express in linear form as shown below:

Ca x 12 x 2 = 0.16 x 36

Divide both side by 12 x 2

Ca = 0.16 x 36/ 12 x 2

Ca = 0.24M

Therefore, the concentration of the acid is 0.24M

6 0
3 years ago
Select all the correct answers.
Lera25 [3.4K]

The correct answers :

It increases with a decrease in the concentration of H₂(g).

It decreases with an increase in the concentration of S₂(g).

It decreases with an increase in the concentration of H₂(g).

<h3 /><h3>Further explanation</h3>

Forward reaction : rate to form product

In equilibrium :

The product decreases ⇒ system will move from left to right(forward reaction)

The product increases ⇒ system will move from right to left(reverse reaction)

6 0
2 years ago
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