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jarptica [38.1K]
3 years ago
14

CHEMISTRY PLEASE HELP ILL GIVE U 20

Chemistry
1 answer:
dybincka [34]3 years ago
5 0
The answer is c hope it helps
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HELP ASAP
WARRIOR [948]
The snake,raccoon, and preying mantis
4 0
3 years ago
Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc
Ber [7]

Answer:

The balanced equation for this reaction will be

                            CH4 + 4F2    →  CF4 + 4HF

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

  • 1 mole of CH4 → 4 mole of 4 mole of fluorine
  • 0.41 mole of methane  →  4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

  • 4 moles of  fluorine →  1 mole of CF4
  • 0.56 mole →  \frac{1}{4} * 0.56 = 0.14mole of CF4
  • 4 moles of fluorine →  4 moles of HF
  • 0.56 mole of fluorine →  0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0
3 years ago
Write a hypothesis that answers the lesson
steposvetlana [31]

Answer:

the reaction will come to a halt and the other reactant will still be present.

7 0
3 years ago
What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0
3 years ago
If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?
Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

  • Volume of Hexane dispensed = 40ml
  • Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

brainly.com/question/4721701

7 0
2 years ago
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