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3 years ago

7

In the chemical equation: NO(g) + ½Cl2(g) → NOCl(g) How many moles of NOCl(g) are in the balanced chemical equation? fill in bla

nk moles NOCl Report your answer as a whole number. Do NOT include units in your answer.

2 answers:

damaskus [11]3 years ago

6 0

<u>Answer:</u> The coefficient of NOCl in the given balanced chemical equation is 1

<u>Explanation:</u>

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

Every balanced chemical equation follows law of conservation of mass

For the given balanced chemical equation:

$NO(g)+\frac{1}{2}Cl_2(g)\rightarrow NOCl(g)$

By Stroichiometry of the reaction:

1 mole of nitrogen monoxide gas reacts with $\frac{1}{2}$ moles of chlorine gas to produce 1 mole of NOCl

Hence, the coefficient of NOCl in the given balanced chemical equation is 1

Mrrafil [7]3 years ago

4 0

Answer:

1

Explanation:

The numbers in front of the formulas (the coefficients) tell you how many moles of a substance are involved in a reaction.

If no coefficient is shown, we assume it is 1.

NOCl has no coefficient, so 1 mol of NOCl is in the balanced equation.

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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:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

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

A substance formed when two<br> or more chemical elements are<br> chemically bonded together?

Ronch [10]

Answer: a piece of pure matter which we can see from our naked eye is termed as a substance

Explanation:

every substance has physical and chemical properties like;

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2 years ago

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3 years ago

PLS HELP What is the mass of hydrogen atoms that is measured at 54 u?

tankabanditka [31]

The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g

<h3>Conversion scale </h3>

1 u = 1.66×10¯²⁴ g

<h3>How to determine the mass of hydrogen atoms </h3>

Mass of Hydrogen (u) = 54 u
Mass of Hydrogen (g) =?

1 u = 1.66×10¯²⁴ g

Therefore

54 u = 54 × 1.66×10¯²⁴ g

54 u = 89.64×10¯²⁴ g

Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g

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dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

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- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

Best regards!

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2 years ago