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2 years ago

How many electrons are there in the following compound?

Chemistry

1 answer:

Marina86 [1]2 years ago

4 0

Answer:

look it up in google it dead in your face m8 i just did

Explanation:

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Describe the range of em radiation

natima [27]

The range for em radiation should be a range of 50 yards hope this help's. <span />

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2 years ago

Which of the following is a statement of Hess's law?

attashe74 [19]

Answer:

Explanation:

the enthalpy of reaction is independent of the reaction path

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2 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

3 years ago

What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?

dusya [7]

When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

6.86 = -㏒ Ka

∴Ka = 1.38 x 10^-7

by using ICE table:

H2PO4- → H+ + HPO4
initial 0.4 m 0 0

change -X +X +X

Equ (0.4-X) X X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

1.38 X 10^-7 = X^2 / (0.4-X) by solving for X

∴X = 2.3x 10^-4

∴[H+] = X = 2.3 x 10^-4

∴PH = -㏒[H+]

= -㏒ (2.3 x 10^-4)
∴PH = 3.6

3 0

2 years ago

What is/are converted into carbon dioxide , water, and energy in aerobic respiration

Lana71 [14]

Cellular respiration involves breakdown of glucose into carbon dioxide and water in presence of oxygen, releasing energy. ... ATP molecule is converted into ADP molecule, whenever energy is needed for any metabolic reaction or activity. The energy stored in it is released to be used in metabolic reaction.

Hope this helps! So sorry if I'm wrong

3 0

3 years ago

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