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kvv77 [185]
3 years ago
5

What is the final step in correctly matching ammunition to a firearm?

Chemistry
1 answer:
bazaltina [42]3 years ago
3 0
The final step in correctly matching ammunition to the fire arm is to MATCH THE INFORMATION ON THE BARREL WITH THE INFORMATION ON THE CARTRIDGE OR THE SHOT SHELL.
Getting the final step right is critical because if this final step is not properly carried out, an explosion can result, which can seriously injure the user and the bystanders or even kill them.
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Isotopes of the same element differ in
horrorfan [7]

They are uniform in the amount of protons and electrons but can differ in the amount of neutrons.

8 0
3 years ago
How many grams of H, are needed to produce 14.19 g of NH3 ?
Simora [160]

mass h

<h3>=</h3>

Explanation:

molar mass is equal to number of mass over moles we take 14.19 over17 molar mass of ammammonium equals 0.8347 incase of moles. 1H=1Nh3 ,so. n.H=0.8347 Mr=2 ,M.H=? Mr=m/n. 0.8347x2=1.669grams

4 0
3 years ago
Why is water displacement an effective way of measuring the volume of an irregular solid?
aksik [14]
Water displamement is effective because there is no way to get an exact and accurate measurement of an irregular solid because you can't measure curved line. Water displamement gives you a proper, exact, and accurate answer.
5 0
4 years ago
Read 2 more answers
2. Which of the changes indicated are oxidations
sergeinik [125]

Answer:

A) Oxidized

B) Reduced

C) Oxidized

D) Oxidized

Explanation:

A) Cu becomes Cu²⁺

oxidation state increased from 0 to +2. It gets oxidize.

B) Sn⁺⁴ becomes Sn²⁺

oxidation state decreased from +4 to +2. It gets reduced.

C) Cr³⁺ becomes Cr⁺⁶

oxidation state increased from +3 to +6. It gets oxidize.

D) Ag becomes Ag⁺

oxidation state increased from 0 to +1. It gets oxidize.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

<em>Consider the following reactions. </em>

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

5 0
4 years ago
A compound has a percent composition of 81.71% C and 18.29% H. What is the empirical formula of this compound?
SCORPION-xisa [38]

Considering the definition of empirical formula, the empirical formula is C₃H₈.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, in first place you know the percent composition:

  • C: 81.71 %
  • H: 18.29%

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 81.71 grams of carbon and 18.29 grams of hydrogen H.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

  • C: \frac{81.71 grams}{12\frac{grams}{mole} }= 6.81 moles
  • H:\frac{18.29 grams}{1\frac{grams}{mole} }= 18.29 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

  • C: \frac{6.81 moles}{6.81 moles}= 1
  • H:\frac{18.29 moles}{6.81 moles}= 2.68 ≅ \frac{8}{3}

To express this relationship in the form of simple integers, it is necessary to multiply by a simple number to achieve this:

  • C: 1×3  =3
  • H:≅ \frac{8}{3}×3= 8

Therefore the C: H mole ratio is 3: 8

Finally, the empirical formula is C₃H₈.

Learn more about empirical formula:

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brainly.com/question/4594902

brainly.com/question/13725914

brainly.com/question/13599051

brainly.com/question/13112542

#SPJ1

4 0
2 years ago
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