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3 years ago

A beaker contains two liquids, water and carbon tetrachloride. which liquid is floating on top?

Chemistry

1 answer:

Fudgin [204]3 years ago

4 0

Water, it is less dense than carbon tetrachloride

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According to the Brönsted-Lowry theory, an acid is ___

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Answer:

is a substance which donates an H^+ or a proton

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3 years ago

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What is the Molar mass of Na3P

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Answer:

99.943071 g/mol

Explanation:

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3 years ago

separating a solid from a liquid by evaporating is called _____ a) filtration b) condensation c) solution d) distillation

tatiyna

D) Distillation is the process used to separate a solid from a liquid through evaporating the liquid..

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4 years ago

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From E=mc²=hv show that wavelength=h/mv

barxatty [35]

Explanation:

To show - from E=mc²=hv show that wavelength=h/mv

Proof -

Given that,

E = mc²

E = hν

By equating both the equations, we get

mc² = hν

Because real particles do not travel at the speed of light, De Broglie submitted velocity ( v ) for the speed of light ( c ).

mv² = hν

Through the equation λ , de Broglie substituted v/λ for ν and arrived at the final expression that relates wavelength and particle with speed.

mv² = hv/λ

⇒λ = hv/mv²

⇒λ = h/mv

Hence showed.

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3 years ago

Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili

Katen [24]

Answer: The value of equilibrium constant for the given reaction is 56.61

Explanation:

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

$\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M$

$\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M$

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

$H_2+I_2\rightleftharpoons 2HI$

Initial: 0.1 0.1

At eqllm: 0.1-x 0.1-x 2x

Evaluating the value of 'x'

$\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$[HI]_{eq}=2x=(2\times 0.079)=0.158M$

$[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M$

$[I_2]_{eq}=0.0210M$

Putting values in above expression, we get:

$K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61$

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0

3 years ago