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3 years ago

Calculate the ph of an aqueous solution that is 0.045 m in hcl(aq) at 25°c

1 answer:

ivann1987 [24]3 years ago

4 0

Remember that HCl is a strong acid will dissociate almost 100%
HCl + H₂O --> H₃O⁺ + Cl⁻

pH = -log[H₃O⁺]
pH = -log(0.045) = 1.35

(The rules for sig figs are that if you take the log of a number with x significant figures, then the result should have x significant decimal places.)

The pH is 1.35

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Alborosie

Answer:

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Activity 2:

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Explanation:

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4 0

3 years ago

Read 2 more answers

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago

Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to

posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

Solution

Combined gas law :

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0

3 years ago

According to reference table adv-10, which reaction will take place spontaneously?

olga_2 [115]

Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V

Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu -> C2+ V = - 0.337V

V= - 0.777V (no spontaneous)

5 0

3 years ago

What allows nutrients to enter Plants and animals are able to be transported

elena55 [62]

I think it’s the cell membrane if you’re talking about animal cells and plant cells.

3 0

3 years ago