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2 years ago

Which phrase describes a valence electron?(1 point)

Chemistry

2 answers:

liraira [26]2 years ago

3 0

Answer: I THINK C

Explanation: It has a 25% chance of being right so I'd wait if I were you just sayin

yulyashka [42]2 years ago

3 0

Answer:

Hello, the answer would be C. An electron in the outer energy level of an atom.

Hope this helped :P

You might be interested in

Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti

Lina20 [59]

Answer: The volume of stock solution needed is 90 mL

Explanation:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=6.0M\\V_1=?mL\\M_2=1.2M\\V_2=450mL$

Putting values in above equation, we get:

$6.0\times V_1=1.2\times 450\\\\V_1=\frac{1.2\times 450}{6.0}=90mL$

Hence, the volume of stock solution needed is 90 mL

6 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr