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photoshop1234 [79]
3 years ago
5

Naming compoundsdoes anyone know the answers for these or at least one of them ?​

Chemistry
2 answers:
Arada [10]3 years ago
6 0

Answer:

KNO3- Potassium Nitrate, KOH- Potassium hydroxide, K2CRO4- Potassium chromate

Explanation:

Vaselesa [24]3 years ago
3 0

Answer:

KNO3=Potassium, nitrogen, oxygen3

KOH= Potassium, oxygen, and hydrogen

K2CRO4=Potassium2, chromium, oxygen4

Compound names

KNO3= Potassium Nitrate

KOH= Potassium hydroxide

K2CrO4= Potassium chromate

Hope it helps

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Cl2 + 2OH− → Cl− + ClO− + H2O
Nataly_w [17]
Cl2(g) -------> Cl-(aq) + ClO-(aq) 

2e- + Cl2(g) -------> 2Cl-(aq) [reduction] 

4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation] 
______________________________________... 
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)
4 0
3 years ago
Need a little help with chemistry :) <br> *if you don’t know, don’t answer*
Blizzard [7]

Answer:

1.33 moles

Explanation:

6 0
3 years ago
Read 2 more answers
In step 2, of the experiment, the procedure uses 3.0M NaOH. However, the student notices that the only solution of NaOH is conce
Luda [366]

Answer:

We need 78.9 mL of the 19.0 M NaOH solution

Explanation:

Step 1: Data given

Molarity of the original NaOH solution = 19.0 M

Molarity of the NaOH solution we want to prepare = 3.0 M

Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

Step 2: Calculate volume of the 19.0 M NaOH solution needed

C1*V1 = C2*V2

⇒with C1 = the concentration of the original NaOH solution = 19.0 M

⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED

⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M

⇒with V2 = the volume  of the NaOH solution we want to prepare = 500 mL = 0.500 L

19.0 M * V2 = 3.0 M * 0.500 L

V2 = (3.0 M * 0.500L) / 19.0 M

V2 = 0.0789 L

We need 0.0789 L

This is 0.0789 * 10^3 mL = 78.9 mL

We need 78.9 mL of the 19.0 M NaOH solution

8 0
3 years ago
What volume, in milliliters, of a 0.997 M KOH solution is needed to neutralize 30.0 mL of 0.0400 M HCl?
deff fn [24]

Answer:

1.2 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

HCl + KOH —> KCl + H₂O

From the balanced equation,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of base, KOH (n₆) = 1

Finally, we shall determine the volume of the base, KOH needed to neutralize the acid, HCl as follow:

Molarity of base, KOH (M₆) = 0.997 M

Volume of acid, HCl (Vₐ) = 30 mL

Molarity of acid, HCl (Mₐ) = 0.0400 M

Volume of base, KOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.04 × 30 / 0.997 × V₆ = 1/1

1.2 / 0.997 × V₆ = 1

Cross multiply

0.997 × V₆ = 1.2

Divide both side by 0.997

V₆ = 1.2 / 0.997

V₆ = 1.2 mL

Thus, the volume of the base, KOH needed to neutralize the acid is 1.2 mL.

7 0
3 years ago
PLS ANSWER FAST THANK UU
andriy [413]
Answer
I think it might be B
Explanation
4 0
3 years ago
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