What is simple displacement reaction???give examples.

What is the ph of a solution that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate. (ka for propanoic acid = 1.3 x 10

max2010maxim [7]

<span>The pH is given by the Henderson - Hasselbalch equation:
              pH = pKa + log([A-]/[HA])
              pH = -log(</span><span>1.3 x 10^-5) + log(0.50/0.40)
              pH = 4.98
The answer to this question is 4.98.
</span>

3 0

4 years ago

What is the pressure inside a 2.0 L bottle filled with 0.25 mol of carbon dioxide gas at 25 °C?

motikmotik

Answer:

3.1atm

Explanation:

Given parameters:

Volume of gas = 2L

Number of moles = 0.25mol

Temperature = 25°C = 25 + 273 = 298K

Unknown:

Pressure of the gas = ?

Solution:

To solve this problem, we use the ideal gas equation.

This is given as;

PV = nRT

P is the pressure

V is the volume

n is the number of moles

R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹

T is the temperature

P = $\frac{nRT}{V}$

Now insert the parameters and solve;

P = $\frac{0.25 x 0.082 x 298}{2}$ = 3.1atm

8 0

3 years ago

The charge on an ion is known as its___number.

a_sh-v [17]

Answer:

oxidation number is correct!! :)

Explanation:

8 0

3 years ago

At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival

evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0

3 years ago

Read 2 more answers

1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH

6 0

3 years ago