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Colt1911 [192]
3 years ago
5

How many moles of Cs are contained in 3 moles of Cs3PO4?

Chemistry
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

\boxed {\boxed {\sf 9 \ moles \ Cs}}

Explanation:

We are given the compound: Cs₃PO₄

According to the formula, 1 mole of cesium phosphate contains 3 moles of cesium, 1 mole of phosphate, and 4 moles of oxygen.

Therefore, there are 3 moles of cesium for 1 mole of cesium phosphate.

\frac {3 \ mol \ Cs}{1 \ mol \ Cs_3PO_4}

We want to calculate the moles of cesium in 3 moles of cesium phosphate, so we multiply the ratio by 3.

3 \ mol \ Cs_3PO_4 *\frac {3 \ mol \ Cs}{1 \ mol \ Cs_3PO_4}

3 *\frac {3 \ mol \ Cs}{1 }= 9 \ mol \ Cs

3 moles of cesium phosphate contains <u>9 moles of cesium.</u>

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The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) and Kp=0.174 at 243°C. Calculate the value of Kc for
Mashcka [7]

Answer:

The Kc of this reaction is 311.97

Explanation:

Step 1: Data given

Kp = 0.174

Temperature = 243 °C

Step 2: The balanced equation

N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 3: Calculate Kc

Kp = Kc *(RT)^Δn

⇒ with Kp = 0.174

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 Latm/Kmol

⇒ with T = the temperature = 243 °C = 516 K

⇒ with Δn = number of moles products - moles reactants  2 – (1 + 3) = -2

0.174 = Kc (0.08206*516)^-2

Kc = 311.97

The Kc of this reaction is 311.97

3 0
3 years ago
Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.
Alborosie

<u>Given:</u>

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

<u>To determine:</u>

The Ka of HNO3

<u>Explanation:</u>

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

            HNO3   ↔    H+      +      NO3-

I            7.50               0                 0

C          -2.48          +2.48              +2.48

E            5.02            2.48              2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

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