QUESTION 10

Which compounds are bases in aqueous solution according to brønsted–lowry theory?.

Fittoniya [83]

Ammonia compounds are bases in aqueous solution according to brønsted–lowry theory.

<h3>What are bases?</h3>

A base is a substance that can neutralize the acid by reacting with hydrogen ions.

Ammonia compounds are based on an aqueous solution according to brønsted–lowry theory because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule which accepts the hydrogen ion is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.

Hence, ammonia compounds are based on an aqueous solution according to brønsted–lowry theory.

Learn more about the bases here:

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3 0

2 years ago

What is the name of B2(SeO4)3

Gnoma [55]

This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.

4 0

3 years ago

Read 2 more answers

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as molarity, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

Find the molar mass (M) of potassium hydroxide.

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

Calculate the moles of potassium hydroxide.

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

Convert the volume of water to litres.

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

Calculate the concentration.

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

what is the pH of a solution prepared from solid, neutral 2-nitrophenol providing a fromal concentration of 0.0353M, given that

REY [17]

Answer:

pH = 4.34

Explanation:

pH= -1/2(logKa) -1/2(log C)

= -1/2( log 5.98*10^-8) -1/2(log 0.0353)

=-1/2(-7.22)-1/2(-1.45)

=3.61+0.725= 4.34

7 0

3 years ago

The volume of a sample of nitrogen is 6.00 liters at 35oC and 0.75 atm. What volume will it occupy at STP?

joja [24]

The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles

n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?

n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles

Agt STP 1 mole= 22.4 L what obout 0.178 moles

= 22.4 x0.178moles/ 1moles =3.98 L( answer C)

3 0

3 years ago