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olga55 [171]
2 years ago
11

How many grams of H2O are in 34.2 grams of NAOH Need ASAP

Chemistry
1 answer:
Brrunno [24]2 years ago
3 0

Answer:

15.438g H2O

Explanation:

First you need to find the reaction equation:

2H2O+2Na=2NaOH + H2

Hydrogen is a diatomic molecule so it will have a subscript of 2 on the right hand side. From there we can balance the reaction.    

Then we can use stoichiometry:

34.2g NaOH * (1 mol NaOH/39.908g NaOH) * (2 mol H2O/2 mol NaOH) * (18.015g H2O/1 mol H20) = 15.438g H2O

It is important that when you use stoichiometry that all your units cancel out until you only have the unit you want.

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You were instructed to add 1.0 mL out of 4.0 mL of an undiluted sample to 99 mL of sterile diluent. Instead, you add the entire
finlep [7]

Answer:

A.) Intended dilution factor : 0.01

B.) Actual dilution factor: 0.039

Explanation:

Dilution factor(DF) is ratio of the volume of the undiluted sample to total volume of diluted solution

A.) What was your intended dilution factor

This would be the the addition 1.0ml from 4.0ml to 99ml of sterile

Volume of sterile diluent = 99ml

Volume of Undiluted sample = 1.0ml

Total volume of diluted solution = 99ml + 1m = 100ml

Dilution factor = Volume of undiluted sample ÷ Total volume of diluted solution

Dilution factor = 1:100

= 1ml ÷ 100ml

= 0.01

B.) What was your actual dilution factor?

This would be the entire addition of 4.0ml to 99ml sterile diluent

Volume of sterile diluent = 99ml

Volume of Undiluted sample = 4.0ml

Total volume of diluted solution = 99ml + 4.0ml = 103ml

Dilution factor = Volume of undiluted sample ÷ Total volume of diluted solution

Dilution factor

= 4:103

= 4ml ÷ 103ml

= 0.039

8 0
3 years ago
Don't know how to solve
faltersainse [42]
Happy Valentines Day! <3
8 0
3 years ago
In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in
Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0
3 years ago
What is a diatom?
mezya [45]

Answer:

a molecule with two of the same element

Explanation:

3 0
3 years ago
2. What is the total mass of Chlorine in ZnCl,?
Vedmedyk [2.9K]

Answer:

35.5

Explanation:

35.5.

relative moleculer mass

6 0
2 years ago
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