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3 years ago

How many grams are in 5.00 * 10^25 atoms of helium

Chemistry

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

dfghgvfdsrftgbnbvcdfvgbhnbvcxsdefrgthbvcxdsfgt

Explanation:

sdfgvfcdsertgyhbvcfxdsrftgh

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20 ptsWhich of the labeled structures breaks down food molecules to make ATP? A. the mitochondria B. the Golgi bodies C. the nuc

vaieri [72.5K]

A i think but i could be wrong

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3 years ago

What pressure is exerted by 0.883 mol N, in a 4.68 L steel container at 197.9 °C?

SpyIntel [72]

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3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

4 years ago

Please answer the “(fill-in)” ones thank you :)

Dovator [93]

Answer:

look

Explanation:

diffculte ou my good i need help too

8 0

3 years ago

What is the oxidation number of each phosphorus atom in H4P2O7

nekit [7.7K]

Answer:

Explanation:

The oxidation number of phosphorus can be obtained as follows:

H4P2O7 = 0

4(+1) + 2P + 7(—2) = 0

4 + 2P —14 = 0

Collect like terms

2P = 14 — 4

2P = 10

Divide both side by 2

P = 10/2

P = +5

The oxidation number of phosphorus is +5

7 0

4 years ago