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nadya68 [22]
3 years ago
7

Which statement describes the blood type of a person with the alleles Ai?

Chemistry
1 answer:
lyudmila [28]3 years ago
5 0

Answer:

d i dont now but im sure my answer is d im not good to write english word

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A compound is found to contain 73.23% xenon name 26.77% oxygen by mass. What is the empirical formula for this compound ?
Luba_88 [7]

The empirical formula is XeO₃.

<u>Explanation:</u>

Assume 100 g of the compound is present. This changes the percents to grams:

Given mass in g:

Xenon = 73.23 g

Oxygen = 26.77 g

We have to convert it to moles.

Xe = 73.23/   131.293 = 0.56 moles

O = 26.77/ 16 = 1.67 moles

Divide by the lowest value, seeking the smallest whole-number ratio:

Xe = 0.56/ 0.56 = 1

O = 1.67/ 0.56 = 2.9 ≈3

So the empirical formula is XeO₃.

6 0
3 years ago
If a pork roast must absorb
Black_prince [1.1K]
Let us  assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2 
heat absorbed by pork = 0.11 x 2217
                                     = 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
                                              = 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
          = 44 x 20.913
          = 920.172g of CO2 emitted 
7 0
3 years ago
Nitrogen-13 has a half-life of about 10 minutes. How much of a 320g sample of N-13 would remain after 40 minutes?
lubasha [3.4K]

Answer:

The half-life of a radioisotope describes the amount of time it takes for said isotope to decay to one-half the original amount present in the sample.

Nitrogen-13, because it has a half-life of ten minutes, will experience two half-lives over the course of the twenty minute period. This means that 25% of the isotope will remain after this.

0.25 x 128mg = 32mg

32mg of Nitrogen-13 will remain after 20 minutes.

5 0
3 years ago
Balance the following half-reaction: (acidic) NO−3(aq) → NO2(aq) Express your answer as a half-reaction. Identify all of the pha
Marrrta [24]

Answer:

2H⁺  + NO₃⁻  + 1e⁻ →  NO₂  + H₂O

Explanation:

NO₃⁻  →  NO₂

In left side, Nitrogen acts with +5 by oxidation number

In right side, the oxidation number is +4

This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.

NO₃⁻  + 1e⁻ →  NO₂

In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.

NO₃⁻  + 1e⁻ →  NO₂  + H₂O

Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side

2H⁺  + NO₃⁻  + 1e⁻ →  NO₂  + H₂O

8 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
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