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vfiekz [6]
3 years ago
13

HS0, + NaOH

Chemistry
1 answer:
tatyana61 [14]3 years ago
7 0

Answer: Synthesis

Explanation: I think is suynthesis

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Calculate the number of mol of helium in a 2.01-L balloon at 29°C and 2.71 atm of pressure. Be sure to answer all parts. Imol
konstantin123 [22]

Answer:

0.220 mol He

Explanation:

Hello,

Based on the ideal gas equation of state:

PV=nRT\\n=\frac{PV}{RT} \\n=\frac{2.71 atm * 2.01 L}{0.082\frac{atm*L}{mol*K}*302K } \\n=0.220 mol

Best regards.

4 0
3 years ago
What is the uncertainty of a 10ml pipet?​
EleoNora [17]

Answer: the certain of a 10ml pipet is listed as, 10.00 0.02,

Explanation: those numbers are close to 4 significant figures as in (10ml) around the amount you're looking for. , hope that makes sense!

3 0
3 years ago
If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79
Helen [10]
1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass. 

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4

3) convert moles of CuSO₄ to moles of Cu

1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu

I did it step-by-step as the explanation but you can do all of this in one step. 

200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu


4 0
4 years ago
What is the difference between weak and strong acids?
Rasek [7]

Answer:

<u>A weak acid</u> is an acid which dissociates partially in a solution to release few hydrogen ions while <u>A strong acid</u> is an acid which completely dissociates in solution to release many hydrogen ions.

8 0
3 years ago
Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −
irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a2

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}

The chemical equation for the second dissociation reaction follows:

H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}

The chemical equation for the third dissociation reaction follows:

HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}

To form a buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a of second dissociation process

To calculate the pK_a, we use the equation:

pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})

pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})

We are given:

pK_a2 = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

[HPO_4^{2-}] = concentration of conjugate base

[H_2PO_4^{-}] = concentration of weak acid

Hence, to calculate the pH of the buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a2

3 0
4 years ago
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