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2 years ago

Using the balanced equation, perform the stoichiometry calculations indicated in a. and b. below

Chemistry

1 answer:

Cloud [144]2 years ago

4 0

Answer:

a) 113g Sr3(PO4)2

b) 1.5g H2

Explanation:

The number of moles of Sr used in the equation

= mass/ molar mass

=65.30/87

= 0.751mol

Since the mole ratio of Sr: Sr3(PO4)2 is 3:1

moles of Sr3(PO4)3 = 1/3*0.751

= 0.25mol

grams of Sr3(PO4)3 is = moles * molar mass

= 0.25 * 452

= 113g

b). Since mole ratio of Sr:H2 is 3:3

moles of H2 = 0.751 * 3/3 ---> 0.751 mol

grams of H2 is = moles * molar mass

= 0.751 * 2

= 1.502g

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Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

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Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

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3 years ago

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Tpy6a [65]

Answer:

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These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

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The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminum is used to block the beta radiation

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Gamma radiations:

Gamma radiations are high energy radiations having no mass.

These radiations are travel at the speed of light.

Gamma radiations can penetrate into the many materials.

These radiations are also used to treat the cancer.

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The lead apron are used by the person when treated with gamma radiations.

Lead shields are also used in the wall, windows and doors of the room where gamma radiations are treated, in-order to protect the surroundings.

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