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GREYUIT [131]
1 year ago
5

an m peak has a relative intensity of 42%. an m 1 peak has a relative intensity of 6.5%. how many carbons are in the molecule?

Chemistry
1 answer:
irakobra [83]1 year ago
5 0

5 carbons are in the molecule. Isotopes are atoms with a constant number of protons but a variable number of neutrons.

<h3>What factors affect an isotope's relative abundance?</h3>

Isotopes are atoms with a constant number of protons but a variable number of neutrons. Atomic masses vary among isotopes.

The proportion of atoms with a particular atomic mass that can be found in a sample of an element that is found naturally is known as the relative abundance of an isotope.

Relative abundance chemistry issues should be solved using the formula below:

(M1)(x) + (M2)(1-x) = M (E)

Issue illustration Determine the relative abundance of the isotopes if the masses of two different nitrogen isotopes are 0.42 amu for carbon-0.42 and 0.065 amu forcarbon-0.065.

Equation x = 0.996 can be found using algebra.

0.42 * 0.996 + 0.065 *0.996 = 0.424794.

5 carbons are in the molecule.  

To learn more about Relative abundance of an isotope refer to:

brainly.com/question/24873591

#SPJ4

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What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C
Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = \frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\

Now use the molar ratio to find how many moles of NaOH would be required to neutralize 5.5537*10^{-3} mol\\ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

5.5537*10^{-3} mol\\ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L

5 0
4 years ago
When a lead storage battery discharges, the concentration of ___.
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3 years ago
22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass
Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃  +  MgBr₂  → Mg(NO₃)₂  +  2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

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0.0662 mol . 148.3 g/ mol = 9.82 g

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3 years ago
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