What is the percent mass/volume of sodium chloride solution made by dissolving 5.0 grams of sodium chloride in 40.0 ml of soluti

zalisa [80]

f

because sun dont orbits

7 0

4 years ago

You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40,

eimsori [14]

According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
= 7.55 - 1.95 = 5.6

5 0

3 years ago

Round the following numbers to the number of significant figures indicated:

Kruka [31]

3.5
9
24.7
8.01
0.006
13(1.25×1012=12.65)

Explanation:

Above we have simply looked the condition that where we put decimal point that gives your needed answer after round off.

For example:

we have to make 2 s.f from 345 so we kept decimal after 3 and we rounded off to make 3.5 from 3.45

Hope you got it

7 0

3 years ago

How many moles is 3.85 x 1025 atoms of gold?

IrinaK [193]

Answer:

64.17 Moles of Au

Explanation:

(atoms and particles are the same)

3.85 x 10 ^25 x (1 mol

/6.02 x 10^23)

3.85 / 6 = .64166

.6416 x 10^2 = 64.166

If you round up the answer you will get 64.17

64.17 moles of Au

4 0

3 years ago

Read 2 more answers

Balance each of the following in BASE: Show work

kupik [55]

Answer : The balanced chemical equation in a basic solution are,

(A) $2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) $2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) $4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) $Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

Explanation :

(A) The given chemical reaction is :

$CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)$

The oxidation-reduction half reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}$