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3 years ago

Please help me with this, thank you so much

Chemistry

1 answer:

Basile [38]3 years ago

7 0

I can’t answer the exact thing, but I know how to do it. One of the nitrogen bases always pairs with a different one, so once you know what pairs with what you just write it down. If you need more help just ask :) tip: one base will only join with a different one. One won’t join to multiple (from what I know as of now)

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Which of the following is not a result when a change to an equilibrium system is applied? (2 points)

Ray Of Light [21]

It should be increasing the rate of the forward reaction will cause a shift to the left because the external stress, which is the increase in rate, will cause the reaction to be unbalanced, and to reach equilibrium it needs to shift to the right.

8 0

3 years ago

Read 2 more answers

A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the

Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

$\displaystyle c = \frac{n}{V}$ ,

where

$c$ is the concentration of the solute,

$n$ is the number of moles of the solute, and

$V$ is the volume of the solution.

$V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}$ .

Acetic (ethanoic) acid:

$\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}$ .

Hydrochloric acid HCl:

$\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}$ .

$\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}$ .

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be $5.00\times 10^{-4}\;\text{M}$ .

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be $-x\;\text{M}$ . $x > 0$ .

$\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}$ .

$\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}$ .

Rewrite as a quadratic equation and solve for $x$ :

$x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)$

$x\approx 0.00111$ .

The pH of a solution depends on its H⁺ concentration.

At equilibrium

$[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}$ .

$\text{pH} = -\log{[\text{H}^{+}]} = 2.79$ .

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3 years ago

What is the mass of silver in 3.4g of AgNO3?

Licemer1 [7]

3.4g AgNO3 x (1 mole AgNO3/169.8732g AgNO3) x (1 mole Ag/1mole AgNO3) x (107.8682g Ag/1 mole Ag) =

2.2 grams Ag 
(2 significant figures

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3 years ago

The constitutional isomer of ethanol, dimethyl ether (CH3OCH3), is a gas at room temperature. Suggest an explanation for this ob

Sedaia [141]

Answer:

Because of its weak intermolecular forces.

Explanation:

Hello there!

In this case, according to the given description, it turns out possible for us to recall the chemical structures of both ethanol and dimethyl ether as follows:

$CH_3CH_2OH\\\\CH_3COCH_3$

Thus, we can see that ethanol have London dispersion forces (C-C bonds), dipole-dipole forces (C-O bonds) and also hydrogen bonds (O-H bonds) which make ethanol a liquid due to the strong hydrogen bonds. On the other hand, we can see that dimethyl ether has just London and dipole forces, which are by far weaker than hydrogen bonding, that makes it unstable when liquid and therefore it tends to vaporize quite readily.

Regards!

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3 years ago

4.053g 4.2 mL = Please help

butalik [34]

Answer: 2

Explanation:

5 0

2 years ago