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3 years ago

15

Which phases describe air density?

1 answer:

Bond [772]3 years ago

7 0

Answer:

Explanation:

we can conclude that phrases described by air density are as follows

equals mass divided by volume.

exerts more pressure as it increases.

measures how tightly molecules are packed

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Which of the following statements is true about taste bud preferences?

Dafna1 [17]

It seems that you have missed the necessary options for us to choose from, but anyway, here is the answer. The statement that is considered true about taste bud preferences would be this: The taste buds are located within the papillae of the tongue. Hope this answers your question.

6 0

3 years ago

Given the incomplete equation representing a reaction:

sergey [27]

Answer : The formula of the missing product is, $OH^_$

Explanation :

The given incomplete equation representing a reaction,

$2Na(s)+2H_2O(l)\rightarrow 2Na^+(aq)+2__(aq)+H_2(g)$

When the sodium metal react with the water then it gives a colorless solution of sodium hydroxide and hydrogen gas. In the solution, sodium hydroxide is present in the form of ions i.e, $Na^+$ and $OH^-$ ions.

The balanced chemical reaction will be,

$2Na(s)+2H_2O(l)\rightarrow 2Na^+(aq)+2OH^-(aq)+H_2(g)$

Therefore, the formula of the missing product is, $OH^_$

3 0

3 years ago

Read 2 more answers

Nataly_w [17]

Answer:

The average atomic mass of oxygen is 15.999u

5 0

3 years ago

A student performs the following experiment.

jenyasd209 [6]

Answer:

yes

Explanation:

not becming

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4 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