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3 years ago

What is the name of this molecule?

Chemistry

1 answer:

Roman55 [17]3 years ago

7 0

I think this very well may be cycloalkanes. <span />

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How many moles of SnCl2 will be produced if 85.3 g of FeCl3 is reacted with an excess of Sn for the following reaction:2 FeCl3(s

Ronch [10]

Answer:

The answer to your question is 0.79 moles of SnCl₂

Explanation:

Data

moles of SnCl₂ = ?

mass of FeCl₃ = 85.3 g

excess Sn

Balanced chemical reaction

2 FeCl₃ + 3 Sn ⇒ 3 SnCl₂ + 2 Fe

Process

1.- Convert the mass of FeCl₃ to moles

Molar mass of FeCl₃ = 56 + (35.5 x 3)

= 56 + 106.5

= 162.5 g

Use proportions to find the moles of FeCl₃

162.5 g -------------------- 1 mol

85.3 g ------------------- x

x = (85.3 x 1) / 162.5

x = 0.525 moles

2.- Find the number of moles SnCl₂

2 moles of FeCl₃ ----------------- 3 moles of SnCl₂

0.525 moles ----------------- x

x = (0.525 x 3) / 2

x = 0.79 moles of SnCl₂

4 0

4 years ago

Which of the following is true of liquids and solids?

gayaneshka [121]

The answer is D. This is because liquids take up the shape of the container they are in, so it is never definite. Where as solids stay the same shape.

3 0

4 years ago

The chemical reaction represented as “AB ⇒ A + B” is a ______

Lana71 [14]

Decomposition reaction

8 0

3 years ago

Read 2 more answers

A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af

larisa86 [58]

Explanation:

The given reaction is as follows.

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Hence, number of moles of NaOH are as follows.

n = $0.05 L \times 0.1 M$

= 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

n = $0.025 L \times 0.1 M$

= 0.0025 mol

According to ICE table,

$HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)$

Initial: 0.005 mol 0.0025 mol 0 0

Change: -0.0025 mol -0.0025 mol +0.0025 mol

Equibm: 0.0025 mol 0 0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

[HA] = $\frac{0.0025 mol}{V}$

[NaA] = $\frac{0.0025 mol}{V}$

$[A^{-}] = [NaA] = \frac{0.0025 mol}{V}$

Now, we will calculate the $pK_{a}$ value as follows.

pH = $pK_{a} + log \frac{A^{-}}{HA}$

$pK_{a} = pH - log \frac{[A^{-}]}{[HA]}$

= $3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}$

= 3.42

Thus, we can conclude that $pK_{a}$ of the weak acid is 3.42.

8 0

3 years ago

In our simulation what did the σ (sigma) stand for?

jenyasd209 [6]

Answer:

The atomic radius corresponds to Sigma

Explanation:

It is the Van Der Waals radius ;)

3 0

3 years ago