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Elza [17]
2 years ago
13

What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

Chemistry
1 answer:
Morgarella [4.7K]2 years ago
5 0

This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5%​ as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

\% H=\frac{m_H*1}{M_{HF}}*100\%

Where m_H stands for the atomic mass of hydrogen and M_{HF} for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%

Learn more about percent compositions: brainly.com/question/12247957

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Which of the following will have the slowest rate of diffusion at a given temperature?
marysya [2.9K]
<h3>Answer:</h3>

Chlorine gas (Cl₂)

<h3>Explanation:</h3>
  • According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
  • Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
  • Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.

In this case we are given gases, CI₂

, H₂,He and Ne.

  • We are required to identify the gas that will diffuse at the slowest rate.
  • In other words we are required to determine the heaviest gas.

Looking at the molar mass of the gases given;

Cl₂- 70.91 g/mol

H₂- 2.02 g/mol

He - 4.00 g/mol

Ne- 20.18 g/mol

Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.

8 0
3 years ago
Niobium-91 has a half-life of 680 years. After 2,040 years, how much niobium-91 will remain from a 300.0-g sample? 3 g 18.75 g 3
Vadim26 [7]
Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
7 0
3 years ago
Read 2 more answers
What is the electron configuration for La (lanthanum)?
MrRissso [65]

Answer:

Lathanum .

Atomic number = 57

Symbol = La

Atomic weight = 138.9

No of energy orbitals = 6

Electronic configuration

[Xe]6s^25d^1

7 0
3 years ago
Given the equation C3H8(g) + O2(g) = CO2(g) + H2O(g) and that the enthalpies of formation for H2O(g) = -241.8 kJ/mol, CO2(g) = -
kari74 [83]

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)  \Delta H=-2220.1kJ/mol

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]

\Delta H_{C_3H_8}=72.4kJ/mol

Therefore, the heat of formation of propane is 72.4 kJ/mol

7 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge=1.6\times 10^{-19}C

1 mole of electrons contain=6.023\times 10^{23} electrons

Thus  1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000Coloumb of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by=\frac{193000}{0.0635}\times 11.5=34952756 Coloumb

Q=I\times t

where Q= quantity of electricity in coloumbs  = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

34952756 C=34.5A\times t

t=1013123sec=281hours

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

8 0
3 years ago
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