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laiz [17]
3 years ago
15

Metals react by losing valence electrons. Which of the alkali metals loses its valence electrons most easily? Explain.

Chemistry
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

lol you have an aot profile pic

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Calculate the pH of a 0.22 M ethylamine solution.
evablogger [386]

Answer:

answer is 12.18

Explanation:

(C2H5NH2, Kb = 5.6 x 10-4.)

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2 years ago
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Looking for 14-16, hispanic, curly hair, genuinely a nice guy
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3 years ago
Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1g of potassium nitrate?
Effectus [21]
In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100

The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams

The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams
4 0
3 years ago
How many moles of ethane (C2H6) would be needed to react with 62.3 grams of oxygen gas?
Vladimir [108]
Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas

Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988

Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol

now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1

1 C2H6 + 1 O2 -> products

this means that for 1 mol of C2H6, 1 mol of O2 has to react with it

Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
5 0
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
3 years ago
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