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Vlada [557]
3 years ago
8

Ultra Clorox bleach is 6.68% sodium hypochlorite in water. How much of each is needed to make an 800.mL bottle?

Chemistry
1 answer:
sweet-ann [11.9K]3 years ago
7 0
If we have 6.68% NaClO, it is the same as saying--> 6.68 grams NaClO= 100 mL of solution. we can use this as a conversion. 

800. mL (6.68 mL/ 100 mL)= 53.4 mL

solution = solute + solvent

solute= NaClO
solvent= H2O

solvent= 800-53.4= 747 mL of H2O

so, we you need 53.4 mL of NaClO and 747 mL of water or 53.4 grams of NaClO and 747 mL of water
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The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni
ValentinkaMS [17]

Answer:

"6.7\times 10^{-4} \ atm" is the right answer.

Explanation:

Given:

Partial pressure of N_2,

= 0.20 atm

Partial pressure of H_2,

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K_p = 1.5\times 10^3 at 400^{\circ} C

As we know,

⇒ K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}

By putting the values, we get

    1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}

        pNH_3^2 = \frac{0.000675}{1.5\times 10^3}

                    =6.7\times 10^{-4} \ atm

                   

3 0
2 years ago
What is the name for this molecule? 3-heptyne 3-heptene 3-hexyne 3-hexene
sweet-ann [11.9K]

The iupac name of the compound will be hex-3-yne or 3-hexyne. By marking number of carbon in the given compound, it was found that triple bond comes at third position and there are total 6 carbon in the compound. So it will have Hex as a prefix and as it contains triple bond so it will have yne as a suffix and as the triple bond is at third position, so it will be hex-3-yne or 3-hexyne.

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3 years ago
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What is the difference between reactants and products?
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Explanation:

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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)]

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

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