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Svetach [21]
4 years ago
5

How many moles are in 4.99x1032 molecules of H2o

Chemistry
1 answer:
bogdanovich [222]4 years ago
5 0

Answer:

n≈8,29×10⁸ moles

Explanation:

Nᴀ=6.02×10²³ (constant)

n=N/Nᴀ=

=  \frac{ {4.99 \times 10}^{32} }{6.02 \times 10 ^{23} }  = 828903654

≈8,29×10⁸ moles

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When a physical change in a sample occurs, which of the following does NOT change?
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I think thee correct answer from the choices listed above is option D. <span>When a physical change in a sample occurs, composition of the sample does not change. It stays the same. Also, the properties of the sample will still be the same. Hope this answers the question.</span>
8 0
3 years ago
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What is the density of a substance that has a mass of 20 g and a volume of<br> 10 mL?
Paha777 [63]

Answer:

That unknown substance is water

Explanation:

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3 years ago
How many milliliters of 0.0896M LiOH are required to titrate 25.0 mL of 0.0759M HBr to the equivalence point?
slavikrds [6]

Answer:

V_{LiOH}=21.8mL

Explanation:

Hello,

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n_{LiOH}=n_{HBr}

That it terms of molarities and volumes we have:

M_{LiOH}V_{LiOH}=M_{HBr}V_{HBr}

Next, solving for the volume of lithium hydroxide we obtain:

V_{LiOH}=\frac{M_{HBr}V_{HBr}}{M_{LiOH}} =\frac{0.0759M*25.0mL}{0.0896M} \\\\V_{LiOH}=21.8mL

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8 0
3 years ago
Wich has a larger percent by mass of sulfur - H2SO or H2S7O8?
Pie

Answer:

Explanation:

H2SO4 let S be x

2(1) + x + 4(-2) = 0

2 + x - 8 = 0

x - 6 = 0

x = 6

For H2S7O8 let S be x

2(1) + 7(x) + 8(-2) = 0

2 + 7x - 16 = 0

7x - 14 = 0

7x = 14

x = 14/7

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5 0
3 years ago
A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?
dem82 [27]

Answer: 1.67\times 10^{-3}

Explanation:

ClCH_2COOH\rightarrow ClCH_2COO^-+H^+

   cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Given:  c = 0.15 M

pH = 1.86

K_a = ?

Putting in the values we get:

Also pH=-log[H^+]

1.86=-log[H^+]

[H^+]=0.01

[H^+]=c\times \alpha

0.01=0.15\times \alpha

\alpha=0.06

As [H^+]=[ClCH_2COO^-]=0.01

K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}

K_a=1.67\times 10^{-3]

Thus the vale of K_a for the acid is 1.67\times 10^{-3}

4 0
3 years ago
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