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fiasKO [112]
3 years ago
7

What volume of 5.00 × 10-3 m hno3 is needed to titrate 80.00 ml of 5.00 × 10-3 m ca(oh)2 to the equivalence point? what volume o

f 5.00 × 10-3 m hno3 is needed to titrate 80.00 ml of 5.00 × 10-3 m ca(oh)2 to the equivalence point? 80.0 ml 160. ml 10.0 ml 40.0 ml?
Chemistry
1 answer:
Dimas [21]3 years ago
4 0

Answer:

= 160 mL

Explanation:

The equation for the reaction is;

Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + 2H2O

Moles of Ca(OH)2 =  0.080 L x 5.00^-3 M

                              = 0.0004 moles

From the reaction; 2 moles of HNO3 reacts with 1 mole of Ca(OH)2

Thus; moles of HNO3 = 0.0004 Moles × 2

                                    = 0.0008 moles

Volume = moles/ molarity

Therefore; volume of HNO3= 0.0008 Moles/ 5 × 10^-3 M

                                              <u>= 0.16 L or 160 mL</u>

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11Alexandr11 [23.1K]

Answer:

[OH-] = 6.17 *10^-10

Explanation:

Step 1: Data given

pOH = 9.21

Step 2: Calculate [OH-]

pOH = -log [OH-] = 9.21

[OH-] = 10^-9.21

[OH-] = 6.17 *10^-10

Step 3: Check if it's correct

pOH + pH = 14

[H+]*[OH-] = 10^-14

pH = 14 - 9.21 = 4.79

[H+] = 10^-4.79

[H+] = 1.62 *10^-5

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Explanation:

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2 years ago
What is the empirical formula of the compound that is 25.3% magnesium and 74.7% chlorine
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Explanation:

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First step: change % to g

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Third step: determine the lowest number and divide everything by it. Of the result, extract whole number only.

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Fourth step: Write each compound with their respective number below.

This empirical formula should be: MgCl_2

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