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3 years ago

What is the % by volume of 50mL of ethylene glycol dissolved in 950mL of H2O?

Chemistry

1 answer:

MariettaO [177]3 years ago

5 0

Answer:

5.0 %

Explanation:

Given data

Volume of ethylene glycol (solute): 50 mL

Volume of water (solvent): 950 mL

Step 1: Calculate the volume of solution

If we assume that the volumes are additive, the volume of the solution is equal to the sum of the volume of the solute and the solvent.

V = 50 mL + 950 mL = 1000 mL

Step 2: Calculate the percent by volume

We will use the following expression.

$\% v/v = \frac{volume\ of\ solute}{volume\ of\ solution} \times 100 \% = \frac{50mL}{1000mL} \times 100 \% = 5.0\%$

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Please help me with number 4 and 5 on my study guide

Shtirlitz [24]

Answer:

help with problem 4

Explanation:

4. protons are equal to atomic numbers. neutrons are atomic mass minus atomic number. electrons are equal to protons.

5. I don't know, sorry.

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2 years ago

Consider the following equilibrium:

Ainat [17]

Answer: The equation which is wrong is $K_p=K_c(RT)^{-5}$

Explanation:

For the given reaction:

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

The expression for $K_c\text{ and }K_p$ is given by:

$K_c=\frac{1}{[O_2]^3}$

$K_p=\frac{1}{[O_2]^3}$

The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.

Relationship between $K_p\text{ and }K_c$ is given by the expression:

$K_p=K_c\times (RT)^{\Delta n_g}$

where,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants

R = gas constant

T= temperature

For the above reaction,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3

Hence, the expression for $K_p$ is:

$K_p=K_c\times (RT)^{-3}$

Therefore, the equation which is wrong is $K_p=K_c(RT)^{-5}$

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3 years ago

Why is the bond between two single hydrogen atoms stable?

Len [333]

B ia the correct one .

Hydrogen only uses the first energy shell, which holds 2 electrons, not 8.

5 0

3 years ago

Read 2 more answers

Please answer with an actual answer and not just put a random word ^-^

WARRIOR [948]

Which eclipse was modeled when the large ball was between the small ball and the light?

The model is a "Lunar Eclipse" (If it was talking about the earth, then yes, it is a lunar eclipse).

Which eclipse was modeled when the small ball was between the large ball and the light?

The model is a "Solar Eclipse".

What does the large ball represent?

The earth.

What does the small ball represent?

The moon.

What does the light source represent?

The sun.

Hope this helps!~ <3

(I can't draw so sorry.)

6 0

3 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

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3 years ago