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Aleksandr [31]
2 years ago
6

When a hydrochloric acid solution is combined with a potassium hydroxide solution, an acid-base reaction occurs. write a balance

d molecular equation for this reaction. Write a net ionic equation for this reaction.
Chemistry
1 answer:
Verizon [17]2 years ago
5 0

Answer:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

Explanation:

Hydrochloric acid is an acid because it releases H⁺ in an aqueous solution.

Potassium hydroxide is a base because it releases OH⁻ in an aqueous solution.

When an acid reacts with a base they form a salt and water. This is a neutralization reaction. The neutralization reaction between hydrochloric acid and potassium hydroxide is:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

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What is the boiling point of a solution of 12.0 g of urea in 165.0 g of water
Nat2105 [25]

Answer:

T = 100.63 °C

Explanation:

To solve this question, we need to know what are we talking about here. In this case, we want to know the boiling point of a solution with Urea in water. This is a colligative property, so, the expression to use to calculate that is the following:

ΔT = m * K / MM * kg water (1)

Where:

ΔT: difference of temperatures (Tb of solution - Tb water)

m: mass of the urea

K: ebulloscopic constant of the water (0.52 ° C / m)

MM: molecular mass of urea

The boiling point of water is 100 °C, we have the mass of the urea, but not the molar mass. The urea has the formula CH₄N₂O, so the molar mass can be calculated using the atomic mass of the elements (I will use a rounded number for this):

MM = 12 + (4*1) + (2*14) + 16 = 60 g/mol

Now, we can calculate the ΔT and then, the boiling point of the solution:

ΔT = 12 * 0.52 / 60 * 0.165

ΔT = 6.24 / 9.9

ΔT = 0.63 °C

the value of ΔT is a difference between the boling point of water and the solution so:

ΔT = Ts - Tw

Ts = ΔT + Tw

Replacing we have:

Ts = 100 + 0.63

<h2>Ts = 100.63 ° C</h2>
8 0
3 years ago
Is MgCO3 a correct chemical formula
spin [16.1K]
Yes you are very correct
5 0
3 years ago
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3241004551 [841]

Answer:

what wrong

Explanation:

5 0
2 years ago
Read 2 more answers
A reactant that does not completely in a chemical reaction_
Andre45 [30]

Answer:

B) The theoretical yield is the amount of product that can be made based on the ... C) The limiting reactant is completely consumed in a chemical reaction. D) The actual yield is the amount of product actually produced by a chemical reaction. ... NH3 + 5 O2 → 4 NO + 6 H2O, if 3.00 mol NH3 were made to react with excess of.

Explanation:

MORE  POWER

4 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
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