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Rudiy27
3 years ago
13

Can someone please answer questions 32 and 33?

Chemistry
1 answer:
Talja [164]3 years ago
5 0
32.) 55.6 mL HCl
33.) 0. 128 M
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A volume of 30.0 mL of a 0.140 M HNO3 solution is titrated with 0.570 M KOH. Calculate the volume of KOH required to reach the e
igomit [66]

Answer:

7.37 mL of KOH

Explanation:

So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,

HNO3 + KOH → KNO3 + H2O

Step 1 : The moles of HNO3 here can be calculated through the given molar mass (  0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.

Mol of NHO3 = 0.140 M * 30 / 1000 L = 0.140 M * 0.03 L = .0042 mol

Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,

0.0042 mol HNO2 * ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH

From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,

Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).

Volume of KOH = 0.0042 mol * ( 1 L / 0.570 mol ) * ( 1000 mL / 1 L ) = 7.37 mL of KOH

5 0
3 years ago
5. What would you expect to happen if the cup of olive oil in question 4a is poured into a container of
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Answer:

shi i dont know

Explanation:

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Mention 4 products formed formed from destructive distillation of coal
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coal gas, gas carbon, coal tar and ammonia liquor,

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3 years ago
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If 7.84 × 107 J of energy is released from a fusion reaction, what amount of mass in kilograms would be lost? Recall that c = 3
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m = 7.84x107/(3x108)2kg  = 7.84x107/9x1016kg = 0.871x10-9 kg = 8.71x10-10 kg





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3 years ago
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Predict the formula of the ionic compound that would result from combining the monatomic ions formed by each pair of elements. M
skad [1K]

Answer:

E - Be and O

A - Mg and N

E - Li and Br

F - Ba and Cl

B - Rb and O

Explanation:

Be and O

Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).

Mg and N

Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).

Li and Br

Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).

Ba and Cl

Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).

Rb and O

Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).

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3 years ago
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