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2 years ago

8

How many oxygen atoms are present in a formula unit of calcium acetate?

1 answer:

julsineya [31]2 years ago

6 0

So we end up with a total of four oxygen atoms for this calcium acetate unit and guys that truly it for this one.

You might be interested in

Two parts that make up a solution are... what

hjlf

Answer: All solutions have two parts: the solute and the solvent. The solute is the substance that dissolves, and the solvent is the substance that dissolves the solute.
Hope that helps (:

5 0

2 years ago

According to the National Institute of Health, ventricular tachycardia is a fast, regular beating of the ventricles that may las

jarptica [38.1K]

Answer:

1 makes the most sense

Explanation:

There is no way blood pressure will be decreased especially if your ventricles are erratically contracting for a long period of time. So 3 and 4 are out. And you said number 2 is wrong so my finally answer is 1.

3 0

2 years ago

How many moles are in 3.01 x 10^23 atoms of zinc?

Lerok [7]

Answer:

<h2>0.5 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.01 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{3.01}{6.02} \\ = 0.5$

We have the final answer as

<h3>0.5 moles</h3>

Hope this helps you

6 0

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

5 0

2 years ago

Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO3 solution hav

Alona [7]

Answer:

pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149

Explanation:

HNO₃ (aqueous) ⇄ H⁺ + NO3⁻

The pH is defined as the negative log of the hydrogen ion concentration

pH = - log [H⁺]

From the question, the hydrogen ion concentration is given as 0.71M, therefore

pH = -log [0.71]

= 0.149

3 0

3 years ago

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