I need the solution of number 2

Determine the [OH−] , Ph, and POH of a solution with a [H+] of 0. 00017 m at 25 °C.

tino4ka555 [31]

Main Answer:

Given

pH = -log[H+]

= -log[0.00017]

= 3.769

We know that

pW = pH + pOH

and pW =14

pOH = 14-pH

=14-3.769

=10.231

According to the definition

pOH = -log[OH-]

10.231 = log[OH-]-1

[OH-]= 5.87 x 10-11

Explanation:

What is pH?

pH is defined as the concentration of H+ ion in the solution. If the pH value is less than 7, then the solution will be acidic. If the pH value is greater than 7, then the solution will be basic.

To know more about pH, please visit:

brainly.com/question/8758541

#SPJ4

4 0

2 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

3 years ago

1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced

koban [17]

Answer:

$\large \boxed{\text{200 g CO}_{{2}}}$

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ: 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol: 1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

$\rm \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}$

2. Calculate the mass of CO₂.

$\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}$

3 0

3 years ago

Based on the activity series, which one of the reactions below will occur (Spontaneous)? A. Mn (s) + NiCl2 (aq) → MnCl2 (aq) + N

kap26 [50]

Answer:

Mn (s) + NiCl2 (aq) → MnCl2 (aq) + Ni

Explanation:

The order of displacement of metals from aqueous solution by another metal is defined by the activity series of metals.

The activity series arranges metals in order of reactivity and increasing electrode potentials. The less negative the electrode potential of a metal is, the less reactive it is and the lower it is found in the activity series.

Nickel has a less negative electrode potential than manganese hence it is displaced from an aqueous solution of its salt by manganese spontaneously.

8 0

3 years ago

Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what could be the reason?

RSB [31]

One reason could be that the water in this experiment, didn’t boil, but instead became a “chemical reaction” from atmospheric pressure which is normal.

While heating two different samples, of
sea level water it decides on a temperature of 102°C and the other boils at 99.2°C. Basically calculating of the percent error.

6 0

3 years ago