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Travka [436]
3 years ago
11

PLEASE HELP ME WITH THIS I WILL GIVE YOU POINTS

Chemistry
1 answer:
eduard3 years ago
7 0

Answer:

1. a

2. c

3. b

4. f

5. d

6. e

Explanation:

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Isaac Newton’s investigations of gravity explained which truth?
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Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
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<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
3 years ago
I can't find the density of this question I need help
zaharov [31]
Density is Mass divided by volume.
4 0
3 years ago
Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
bija089 [108]

<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=K_1\times K_2

We are given:  

K_1=K_b

K_2=\frac{1}{K_w}

Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

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