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2 years ago

10

Magnesium has three common isotopes.

1 answer:

sladkih [1.3K]2 years ago

7 0

Answer:

Explanation:

percentage abundance of third isotope = 100 - ( 78.900 + 10.009)

= 11.091 %

Atomic mass

24.1687 x .789 + 25.4830 x .10009 + 24.305 x .11091

19.069 + 2.5506 + 2.69566

= 24.3153 amu

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If you see trapped bubbles ,you have to tap the cover slip using a

trapecia [35]

Answer:

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2 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

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2 years ago

A microwave oven produces energy waves with wavelengths that are

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Longer than visable light

7 0

3 years ago

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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

d1i1m1o1n [39]

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 0.70

Putting values in above equation, we get:

$0.70=-\log[H^+]$

$[H^+]=10^{-0.70}=0.199M$

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated 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 concentrated nitric acid solution

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

We are given:

$M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL$

Putting values in above equation, we get:

$7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL$

Hence, the volume of concentrated solution required is 9.95 mL

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2 years ago

How many grams are in 6.00 moles of NaCl?

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Is it 350.4, I assume?

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2 years ago

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