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3 years ago

9

Use the information and table to answer the following question.

2 answers:

Feliz [49]3 years ago

4 0

Answer:

Explanation:

are you doing your mid term?well i got c

bixtya [17]3 years ago

4 0

Answer:

D

Explanation:

Compounds 1 and 3 are lonic.Because ionic compounds are composed of oppositely charged ions that are attracted to one another,

they require more energy to melt than covalent bonds. just usign he process of elimination you get rid of A and B becuase covalant compound contaIN same negatively charged particles that contain requrie less energy to melt and since two and four have low melting point its d

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Cómo escribir el símbolo Theta

Pani-rosa [81]

Para escribir el símbolo theta, presione 03B8 + Alt + X.

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3 years ago

Read 2 more answers

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

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3 years ago

Which statement best explains what causes the phase change from solid ice to liquid water?

Fofino [41]

It would be c because it has to get hot so it is absorbing the heat not releasing it

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3 years ago

The vapor pressure of benzene is 73.03 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectr

alexira [117]

Answer:

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

Explanation:

The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

Where:

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of the solution

$n_1$ = Number of moles of solvent

$n_2$ = Number of moles of solute

$p_o = 73.03 mmHg$

$p_s= 71.61 mmHg$

$n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol$

$\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}$

$n_2=0.05499 mol$

Mass of 0.05499 moles of estrogen :

= 0.05499 mol × 272.4 g/mol = 14.9802 g

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

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3 years ago

What is the chemical formula for K and P?

SCORPION-xisa [38]

Answer:

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3 years ago