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Morgarella [4.7K]
2 years ago
14

What is the concentration of a solution made with 0.150 moles of KOH in 400.0 mL of solution?

Chemistry
1 answer:
otez555 [7]2 years ago
6 0

Answer:

concentration = \frac{0.15}{0.4}=0.375 mol/L

Explanation:

Concentration: i is defined as the mole per litre.

concentration = \frac{mole}{volume in L}

mole=0.15

volume=400 ml=0.4 litre

concentration = \frac{0.15}{0.4}=0.375 mol/L

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A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?
REY [17]

Answer:

Sn_2O

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1

Therefore, the empirical formula is:

Sn_2O

Best regards.

8 0
2 years ago
A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.7 nm.93.7 nm. What is the final state of the hy
sammy [17]

Answer:

5

Explanation:

Given that the formula is;

1/λ= R(1/nf^2 - 1/ni^2)

λ = 93.7 nm or 93.7 * 10^-9 m

R= 1.097 * 10^7 m-1

nf = ?

ni = 1

From;

ΔE = hc/λ

ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9

ΔE = 21 * 10^-19 J

ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)

-0.963 = (1/nf^2 - 1)

-0.963 + 1 = 1/nf^2

0.037 = 1/nf^2

nf^2 = (0.037)^-1

nf^2 = 27

nf = 5

7 0
2 years ago
A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeyer flask and diluted with 20 mL of 1 M aqueous su
Minchanka [31]
Following reaction is involved in present system:

2KMnO4  +  5H2O2 +  3H2SO4 →  2MnSO4    +     K2SO4  + 5O2  +   8H2O

From the above balance reaction, it can be seen that 2 moles of KMnO4 is consumed for every 5 moles of H2O2.

 Now, percent by mass of hydrogen peroxide in the original solution can be estimated as follows:
percent by mass = \frac{\text{mass of H2O2(g)}}{\text(volume of H2SO4(ml))}X 100
∴percent by mass = \frac{\text{1}}{\text(25)}X 100
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5 0
2 years ago
What are the answers to these questions
Virty [35]

Answer:

Explanation:

So basically it just asking you question about that surtain subject .

5 0
2 years ago
Compound a on ozonolysis yields acetophenone and propanal. what is the structure of compound a? 1-phenyl-1-hexene 1-phenyl-2-pen
astra-53 [7]
Answer:
            2-Phenyl-2-Penetene on ozonolysis <span>yields acetophenone and propanal.

Explanation:
                   The tricky way to solve such questions is to simply break the double bond in alkene and place oxygen atom at each broken half double bond making it carbonyl group. The reaction of given statement is as follow,</span>

6 0
2 years ago
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