Plssss help i’m taking a test xxx

A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/m

Ahat [919]

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

$Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70}$ = 0.08004

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,

$Moles= \frac{8.70\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.2175\ mol$

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{100\ g}{18.0153\ g/mol}$

$Moles\ of\ water= 5.5508\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}$

$Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377$

3 0

3 years ago

The oxidation numbers of nitrogen in NH4+, N2O5, and NaNO3 are, respectively:

Andrew [12]

-III I
NH₄⁺

V -II
N₂O₅

I V -II
NaNO₃
-----------------------

-3 +5 +5

6 0

3 years ago

Read 2 more answers

Which structure is the sun located

lana [24]

It is located in the inner rim of the Orion arm!

3 0

3 years ago

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 25.0 g of aluminum? Express your answe

m_a_m_a [10]

Answer:

A. From 25,0 g of aluminium you will produce 0,927 moles of aluminium chloride

B. From 30,0 g of chlorine gas you will produce: 0,282 moles of aluminium chloride

Explanation:

a. The reaction that you have in this problem is:

3 Cl₂ + 2 Al → 2 AlCl₃

25,0 g of aluminium are:

25,0 ₓ $\frac{1mole}{26,98g}$ = 0,927 moles

As Al:AlCl₃ ratio is 1, from 25,0 g of aluminium you will produce: <em>0,927 moles of aluminium chloride</em>

b. 30,0 g of chlorine gas are:

30,0 ₓ $\frac{1mole}{70,90g}$ = 0,423 moles of Cl₂

As Cl₂:AlCl₃ ratio is ³/₂:

0,423 moles of Cl₂ ₓ $\frac{2 molesAlCl_3}{3 molesCl_2}$ = <em>0,282 moles of aluminium chloride</em>

Thus, from 30,0 g of chlorine gas you will produce: <em>0,282 moles of aluminium chloride</em>

I hope it helps!

5 0

4 years ago

How much hcl must be added to a liter of buffer that is 1. 5 m in acetic acid and 0. 70 m in sodium acetate to result in a buffe

yaroslaw [1]

For a liter of a buffer that is 1. 5 m in acetic acid and 0. 70 m in sodium acetate, and with a result in a buffer pH of 4.01, the moles of HCl required is mathematically given as x=0.313mol

<h3>What is the result in a buffer ph of 4. 01?</h3>

Generally, the equation for the pH is mathematically given as

pH=pKa+logsalt/acid

Generally, the equation for the Chemical Reaction is mathematically given as

HCl+CH3CooNa---->NaCl+CH3OOH

Therefore

4.01=4.74+log\frac{0.8-x}{1.3+x}

-0.52=log\frac{0.8-x}{1.3+x}

x=0.313mol

In conclusion, The moles of HCl required is

x=0.313mol