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4 years ago

Look at the following hypothetical reaction: A+3B-->C+3D

Chemistry

2 answers:

jonny [76]4 years ago

7 0

Not sure why you reposted your question, the answer is D as explained in my other answer.

Elis [28]4 years ago

5 0

Answer: d) A:B

Explanation: Limiting reagent is the reagent which limits the formation of products.

Excess reagent is the reagent which is present in excess in the reaction.

$A+3B\rightarrow C+3D$

a) A:4B

As seen from the given equation, 1 mole of A reacts with 3 moles of B, thus (4-3)= 1 mole of B will remain and thus A is the limiting reagent and B is the excess reagent.

b) A:5B

As seen from the given equation, 1 mole of A reacts with 3 moles of B, thus (5-3)= 2 moles of B will remain and thus A is the limiting reagent and B is the excess reagent.

c) 2A:7B

As seen from the given equation, 1 mole of A reacts with 3 moles of B, thus 2 moles of A will react with 6 moles of B. Thus (7-6)= 1 mole of B will remain and is the excess reagent.

d) A: B

As seen from the given equation, 3 moles of B react with 1 mole of A, thus 1 mole of B will react with 0.33 moles of A and (1 - 0.33) =0.66 moles of A will remain as excess reagent. B is the limiting reagent.

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What are the 4 different types of bonds and how are they formed?

miv72 [106K]

There are four types of chemical bonds essential for life to exist: Ionic Bonds, Covalent Bonds, Hydrogen Bonds, and van der Waals interactions. We need all of these different kinds of bonds to play various roles in biochemical interactions. These bonds vary in their strengths.

To play a variety of roles in biochemical interactions, we require all of these diverse sorts of linkages. The tensile strength of these linkages varies. In chemistry, we consider the range of strengths between ionic and covalent bonds to be overlapping. This indicates that in water, ionic bonds usually dissociate. As a result, we shall consider these bonds from strongest to weakest in the following order:

Covalent is followed by ionic, hydrogen, and van der Waals.

To know more about 4 different types of bonds, visit;

brainly.com/question/17401243

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8 0

1 year ago

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:

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

5 0

3 years ago

Explain how you would make 450 mL of a .250M NaOH solution?

Neporo4naja [7]

Answer:

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

Explanation:

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of NaOH = x

Molar mass of NaOH = 40 g/mol

Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)

Molarity of the solution of NaOH = 0.250 M

$Molarity=\frac{1.248 g}{26 g/mol\times 0.9102254 L}=0.528 mol/L$

$0.250 M=\frac{x}{40 g/mol\times 0.450 L}$

Solving for x:

x = 4.5 g

Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.

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4 years ago

What is the function of this instrument?

chubhunter [2.5K]

To measure weight of a item

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4 years ago

Which of the following is an alkaline earth metal?

svetlana [45]

Magnesium is the answerr

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3 years ago