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harkovskaia [24]
3 years ago
9

You are using a beaker to measure volume. the beakers graduation begin at 20ml and in indicate every 10ml up to 80 ml. the botto

m of the meniscus is about three fourths of the way between 40ml mark and the 50 ml mark. you should report this number as:
a)40ml
b)47ml
c)50ml
d)47.0 ml
Chemistry
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

b) 47ml

Explanation:

The beaker would look like this:

60 mL

50 mL

-------  The bottom of the meniscus is here

40 mL

30 mL

So <u>we have between 40 and 50 mL</u>. There are 10 mL of "distance" between those two values, and three fourths of that difference is (10 * 3/4) 7.5 mL.

So <em>the volume is 47.5 mL</em>, but because the measurement instrument (the beaker) does not handle decimals, the value we should report is 47 mL.

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PLS HELPPP<br> why is CH3OH more polar than CH3CH2OH
dusya [7]

Answer:

Explanation:

Polarity is about differencens in electronegativity. CH bonds have around the same electronegativity value so a CH bond is nonpolar. The more CH bonds there are in a molecule, the more nonpolar it is. Since CH3CH2OH has more carbon-hydrogen bonds than CH3OH, it is more nonpolar. With the same reasoning, since CH3OH has less CH bonds, it's more polar.

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2 years ago
Help please I really really need it now ​
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2 years ago
in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?
dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of C_3H_5N_3O_9

\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol

Now we have to calculate the moles of CO_2,O_2,N_2\text{ and }H_2O

The balanced chemical reaction is:

4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)

From the balanced chemical reaction we conclude that,

As, 4 moles of C_3H_5N_3O_9 react to give 12 moles of CO_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{12}{4}=3 moles of CO_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 1 moles of O_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{1}{4}=0.25 moles of O_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 6 moles of N_2

So, 1 moles of C_3H_5N_3O_9 react to give \frac{6}{4}=1.5 moles of N_2

and,

As, 4 moles of C_3H_5N_3O_9 react to give 10 moles of H_2O

So, 1 moles of C_3H_5N_3O_9 react to give \frac{10}{4}=2.5 moles of H_2O

Now we have to calculate the mole fraction of water.

\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}

\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

p_{H_2O}=X_{H_2O}\times p_T

where,

p_{H_2O} = partial pressure of water vapor gas  = ?

p_T = total pressure of gas  = 58 atm

X_{H_2O} = mole fraction of water vapor gas  = 0.345

Now put all the given values in the above formula, we get:

p_{H_2O}=X_{H_2O}\times p_T

p_{H_2O}=0.345\times 58atm=20.01atm

Therefore, the partial pressure of the water vapor is, 20.01 atm

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3 years ago
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5 bromo 1.1.1 -trichloro -5 ethyl 2.3 dimethyl heptane

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3 years ago
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