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3 years ago

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Making solutions is an extremely important component to real-life chemistry. If you make 3.00 L of a solution using 90.0 g of so

dium hydroxide, what is the final concentration?

1 answer:

kumpel [21]3 years ago

5 0

Answer:

Final concentration of NaOH = 0.75 M

Explanation:

For $NaOH$ :-

Given mass = 90.0 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{90.0\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 2.2502\ mol$

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Where, Volume must be in Liter.

It is denoted by M.

Given, Volume = 3.00 L

So,

$Molarity=\frac{2.2502\ mol}{3.00\ L}=0.75\ M$

<u>Final concentration of NaOH = 0.75 M</u>

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Refer to the electron distribution to answer 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹. What would be the period or series number of

Salsk061 [2.6K]

Answer:

C.5

Explanation:

A number of electrons present in valence shell of penultimate Shell represents the group of elements.

For s block elements: no.of group=number of valence shell electron.

p block elements: no. of group= 2 + 10 + number of valence electrons.

d block elements: no. of group= number(n-1)d electrons + number of electrons in nth shell.

Here, the differential electron is in p orbital hence, it belongs to p block

No. of group= 2+ 10 + 3=15 i.e 15th group or VA group.

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2 years ago

Read 2 more answers

What element has 31 protons and 37 neutrons?

Vlada [557]

The answer is going to be Rubidium. hope that helped

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2 years ago

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

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

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

5 0

2 years ago

The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 3.69 × 1017 at 25°c. calculate e o for a cell m

frutty [35]

We are going to use this formula:

E° = 0.0592/ n * ㏒K

when E° is the standard state cell potential

and n is number of moles of electrons transferred in the balanced equation

for the reaction of the cell.

Sr(s) + Mg2+ (aq) ↔ Sr2+(aq) + Mg(s)

we can get it by using the given balanced equation, we here have 2 electrons

transferred so, n = 2

and K the equilibrium constant = 3.69 x 10^17

so, by substitution:

∴ E° = 0.0592 / 2 * ㏒ (3.69 x 10^17)

= 0.52 V

8 0

3 years ago

What kind of inhibitors are transition state analogs usually classified as? uncompetitive inhibitors competitive inhibitors nonc

Masteriza [31]

<h2>Competitive inhibitors.</h2>

Explanation:

▪ Transition state analogs can be used as inhibitors in enzyme-catalyzed reactions which are done by blocking the active site of the enzyme.

▪ A transition state analog is somewhat the same as that of the transition state.

▪These are better inhibitors than the substrate analogs in competitive inhibition, the reason is that they bind tighter to the enzyme rather than the substrate.

▪ Thus, they are classified as competitive inhibitors.

3 0

3 years ago