How many moles of gas are in 4.4 mL at 1.2 atm?

Which of these is a biomass energy source the sun sugarcane natutal gas water

mixer [17]

I believe the answer is sugarcane

4 0

3 years ago

Can anyone think of some interesting facts about hydrogen? I know that there are plenty, I'm just drawing a blank right now. I'l

podryga [215]

hydrogen is the lightest element in the periodic table

7 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

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Consider a solution containing .100 M fluoride ions and .126M hydrogen fluoride. The concentration of fluoride ions after the ad

S_A_V [24]

Given:

Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M

Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution

Assume: 50:50 ratio of fluoride ions and HF

12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-

5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles

Assume: Volume additive

Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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3 0

3 years ago

what type of science allows us to alter the DNA of a corn plant so that it will contain more protein?

fiasKO [112]

Answer:

biotechnology

Explanation:

hope this helps have a good day

4 0

3 years ago

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