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postnew [5]
3 years ago
7

Human genes can combine in about _________________ different combinations.

Chemistry
1 answer:
Law Incorporation [45]3 years ago
6 0
70,368,744,177,664
i hope that helps!
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Entropy measures _____.<br> a. disorder<br> b. energy<br> c. heat transferred<br> d. force
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A. disorder

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(MC)Which element’s valence electrons are located in 4s24p3?
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Arsenic's valence electrons are there. 
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12 + NaF --&gt; Nal + F2 What is the coefficient of Nal when this equation is<br> balanced? *
e-lub [12.9K]

Answer:

Explanation:

NaL

Difluorine - F2

Molar Mass Bond Polarity F₂ Fluorine Gas Fluorine

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Sodium Fluoride - NaF

Molar Mass Bond Polarity Oxidation State Floridine Sodium Monofluoride Naf

L2

8 0
3 years ago
25 points and ill mark brainlist if i can
MrMuchimi

Hiya!

I believe the answer is...

C.

I hope this is right!

Sokka

3 0
3 years ago
Read 2 more answers
A 20.00 mL Ba(OH)_2 solution of unknown concentration was neutralized by the addition of 47.67 mL of a 0.1236 M HCl solution. Wr
dalvyx [7]

Answer:

- Balanced chemical reaction:

2HCl(aq) + Ba(OH)_2(aq) \rightarrow BaCl_2(aq) + 2H_2O(l)

- Concentration of the initial barium hydroxide solution:

M_{Ba(OH)_2}=0.1473M

- Concentrations of barium and chloride ions:

M_{Ba^{2+}}=0.04353M\\M_{Cl^-}=0.08707M

Explanation:

Hello,

In this case, the balanced chemical reaction with the proper physical states is:

2HCl(aq) + Ba(OH)_2(aq) \rightarrow BaCl_2(aq) + 2H_2O(l)

Moreover, it is convenient to perform the titration analysis in terms of normalities, therefore, HCl's normality becomes equal to its molarity as it is monoprotic (only one hydrogen in its formula), thus, we find barium hydroxide's normality:

N_{Ba(OH)_2}=\frac{N_{HCl}V_{HCl}}{V_{Ba(OH)_2}}=\frac{0.1236N*47.67mL}{20.00mL}=0.2946N

Hence, for the molarity, since two hydroxiles are in the barium hydroxide, 2 grams-equivalents are 1 mole, thus, its molarity is:

M_{Ba(OH)_2}=0.2946\frac{eq-g}{L}*\frac{1mol}{2eq-g}=0.1473\frac{mol}{L}=0.1473M

Then, after mixing and carry out the neutralization reaction the resulting volume is:

V_{final}=20.00mL+47.67mL=67.67mL=0.067L

In addition, the moles of both barium and chloride ions are:

n_{Ba^{2+}}=0.02L*0.1473\frac{molBa(OH)_2}{L} *\frac{1molBa^{2+}}{1molBa(OH)_2} \\n_{Ba^{2+}}=2.946x10^{-3}mol Ba^{2+}\\\\n_{Cl^-}=0.04767L*0.1236\frac{molHCl}{L} *\frac{1molCl^-}{1molHCl} \\n_{Cl^-}=5.892x10^{-3}mol Cl^-

Finally, the concentrations of such ions are:

M_{Ba^{2+}}=\frac{2.946x10^{-3}molBa^{2+}}{0.06767L}= 0.04353M\\M_{Cl^-}=\frac{5.892x10^{-3}molCl^-}{0.06767L}= 0.08707M

Best regards.

5 0
3 years ago
Read 2 more answers
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