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3 years ago

I paid for help here but my questions are not being answered, I think I should delete this and try chegg

Chemistry

2 answers:

luda_lava [24]3 years ago

6 0

Answer:

I'm sorry you feel that way.

Explanation:

What questions do you need answers for?

DENIUS [597]3 years ago

6 0

Answer:

Maybe you should might help

Explanation:

hope I helped MARK ME BRAINELESS IF I HELPED HAVE A GREAT DAY <3

AT LEAST I REPLIED LOL

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What is the shape of name of Co2

Gennadij [26K]

Answer:

The shape is called a Linear shape

The molecule is called Carbon dioxide molecule.

Explanation:

The shape is linear because of the strong repulsive force between the lone pairs

4 0

3 years ago

Transferring or sharing electrons between atoms forms

3241004551 [841]

It forms something called a bond.

4 0

3 years ago

Does anyone have geometry on edg ??!!!

AfilCa [17]

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3 0

3 years ago

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, $(NH_{2})_{2}CO$ .

7 0

3 years ago

A 10.0-milliliter sample of NaOH(aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH(aq)?

Ludmilka [50]

$\frac{10 mL}{1000 mL} = 0.01 L$

$\frac{40mL}{1000mL} =0.04L$

$M _{1} V_{1}=M_{2}V_{2}$

$M_{1}(0.01)=(0.50)(0.04)$

$M_{1}(0.01)=0.02$

$M_{1}= \frac{0.02}{0.01}$

$M_{1}=2$

3 0

4 years ago