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vesna_86 [32]
2 years ago
7

2. _Sn+ _NaOH→_NaSnO2 + _H2 3._H2SO4+ _NaCN » _HCN + _Na2SO4

Chemistry
1 answer:
Slav-nsk [51]2 years ago
6 0

Answer:

2. <u>2</u>Sn+ <u>2</u>NaOH→<u>2</u>NaSnO2 + _H2

3._H2SO4+ <u>2</u>NaCN » <u>2</u>HCN + _Na2SO4

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Measurements show that unknown x compound has the following composition:
Fudgin [204]

The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.

<h3>What is an empirical formula?</h3>

It is the minimum ratio between the elements that form a compound.

  • Step 1: Divide each percentage by the molar mass of the element.

C: 18.1/12.01 = 1.51

H: 2.27/1.01 = 2.25

Cl: 79.8/35.45 = 2.25

  • Step 2: Divide all the numbers by the smallest one.

C; 1.51/1.51 = 1

H: 2.25/1.51 ≈ 1.5

Cl: 2.25/1.51 ≈ 1.5

  • Step 3: Multiply all the numbers by 2 so all of them are whole.

C: 1 × 2 = 2

H: 1.5 × 2 = 3

Cl: 1.5 × 2 = 3

The empirical formula is C₂H₃Cl₃.

The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.

Learn more about empirical formula here: brainly.com/question/1603500

#SPJ1

7 0
2 years ago
PLEASE HELP ME !!!!!!!
love history [14]

Answer:

it would be 5,045

Explanation:

because it is closer to 5,000. pls correct me if wrong

5 0
2 years ago
50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol
boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

5 0
3 years ago
Calculate the concentration of OH- in a solution that contains 3.9 x 10-4 M H3O+ at 25°C. Identify the solution as acidic, basic
aalyn [17]

Answer:

[OH⁻] = 2,6x10⁻¹¹

Acidic

Explanation:

The kw in water is:

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

kw = [OH⁻] [H₃O⁺] = 1,00x10⁻¹⁴

If concentracion of H₃O⁺ is 3,9x10⁻⁴M:

[OH⁻] [3,9x10⁻⁴M] = 1,00x10⁻¹⁴

<em>[OH⁻] = 2,6x10⁻¹¹</em>

pH is defined as - log[H₃O⁺]. If pH>7,0 the solution is basic, if pH<7,0 solution is acidic, if pH=7,0 solution is neutral.

In this problem,

pH = - log [3,9x10⁻⁴M] = <em>3,4</em>

As pH is < 7.0, the solution is <em>acidic</em>

<em></em>

I hope it helps!

8 0
3 years ago
What does temperature indirectly measure? 1)Sound 2)Thermometers 3)Chemical Reactions 4)Heat​
attashe74 [19]

Answer:

Heat

Explanation:

Temperature is the average thermal energy (or heat) in a substance. Sound isn't really related, thermometers measure temperature, and chemical reactions can produce heat, but none of those are measured by temperature.

3 0
2 years ago
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