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3 years ago

HELP PLEASE !!!! DUE NOW!!

Chemistry

1 answer:

ratelena [41]3 years ago

6 0

Answer:

The ratio of the products to reactants remains caonstant over time

Explanation:

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Write L values for the following energy levels. What orbital shapes exist in each of

den301095 [7]

Answer:

n l m

��

1 0 0 1s 1 2 2

��

2 0 0 2s 1 2

2 1 1,0,-1 2p 3 6 8

��

3 0 0 3s 1 2

3 1 1,0,-1 3p 3 6

3 2 2,1,0,-1,-2 3d 5 10 18

��

4 0 0 4s 1 2

4 1 1,0,-1 4p 3 6

4 2 2,1,0,-1,-2 4d 5 10

4 3 3,2,1,0,-1,-2,-3 4f 7 14 32

Explanation:

4 0

2 years ago

Select the most likely product for this reaction:

Rama09 [41]

Answer:

Explanation:

just took it on EDG 2020

6 0

3 years ago

Read 2 more answers

Which of the following is the correct balanced equation for the reaction in which methane (CH4) burns in atmospheric oxygen (0₂)

Talja [164]

Answer: $\text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}$

Explanation:

The unbalanced equation is

$\text{CH}_{4}+\text{O}_{2} \longrightarrow \text{CO}_{2}+\text{H}_{2}\text{O}$

Balancing this equation, we get:

$\boxed{\text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}}$

8 0

2 years ago

Can someone please answer this question please I really need to know

Zarrin [17]

Answer: aging?

Explanation: sorry, i’m not too sure, but that would be my best guess.

8 0

3 years ago

Unit: Chemical Quantities

Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

3 0

4 years ago