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3 years ago

Hydrofluoric acid (HF) can be prepared according to the following equation:

Chemistry

2 answers:

White raven [17]3 years ago

7 0

Answer:

The Hf is 34.78

Explanation:

first put it to moles with the grams times 1mol/ the atomic mass of the element or compound

Artemon [7]3 years ago

3 0

Given the balanced equation:

( Reaction type : double replacement)

CaF2 + H2SO4 → CaSO4 + 2HFI

We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.

This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.

For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.

For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.

Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.

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Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation

nikdorinn [45]

Answer: The isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

Explanation:

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is $_{56}^{139}\textrm{Ba}$

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is $_{38}^{89}\textrm{Sr}$

Hence, the isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

7 0

3 years ago

True or false: Fungi are able to carry out photosynthesis, similar to plants

QveST [7]

Answer:

false

Explanation:

7 0

2 years ago

Read 2 more answers

If I have 8 moles of Iron atoms in the products, how many moles of Iron atoms did I start with?

Vika [28.1K]

Answer: You started with 8

Explanation: the amount of products is equal to the amount of reactants

6 0

3 years ago

Read 2 more answers

17. Which of the following is NOT an empirical formula? *

Aloiza [94]

Answer: 17) d. $C_2H_6$

18. c. The empirical formula of a compound can be twice the molecular formula.

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

The empirical mass can be calculated from empirical formula and molar mass must be known.

17. Thus the empirical formula of $C_2H_6$ should be $CH_3$

18. The molecular formula will either be same as empirical formula or is a whole number multiple of empirical formula. Thus the empirical formula of a compound can never be twice the molecular formula.

3 0

3 years ago

Read 2 more answers

The energy released from 1 gram of uranium is more than 1 million times greater than the energy released from 3 grams of coal. T

SVEN [57.7K]

The energy released from 1 gram of uranium is more than 1 million times greater than the energy released from 3 grams of coal is True.

Explanation:

Nuclear Fission is the process in which splitting of a nucleus takes place that releases free neutrons and lighter nuclei. The fission of heavy elements like "Uranium is highly exothermic" and releases "200 million eV" compared to the energy that is released by burning coal which gives a few eV.

In the given example, it is obvious that the energy released from 1 gram of uranium is more than that of the energy released from 3 grams of coal because the amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only $\frac{1}{10}^{th}$ part of the original nuclei is converted to energy.

5 0

3 years ago