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netineya [11]
3 years ago
14

An Arrhenius base yields which ion as the only negative ion in an aqueous solution?

Chemistry
1 answer:
ser-zykov [4K]3 years ago
7 0

Arrhenius bases are substances which produces hydroxide ions in solution. The <span>only negative ion in an aqueous solution is the hydroxide ion. The answer is number 4.</span>

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Cloud [144]

Answer:

mL

Explanation:

milliliters

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Why is Galileo Galilei impotant?
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Answer:

He provided a number of scientific insights that laid the foundation for future scientists. He also improved telescope that helped further the understanding of the world and universe.

Explanation:

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3 years ago
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Identify three nonmetal atoms that, when bonded covalently, have an argon electron configuration.
Advocard [28]

Answer:

Phosphorus

Sulphur

And chlorine are non metals when bonded covalently gain electronic configuration of argon

Like PCl3

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Explanation:

8 0
3 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

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2 years ago
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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,

In this case, during titration at the equivalence point, we find that the moles of the base equals the moles of the acid:

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

Best regards.

8 0
2 years ago
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