Weak acids

A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62

vesna_86 [32]

Answer: The concentration of $H_2SO_4$ is 0.234 M

Explanation:

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity $H_2SO_4$ = 2

$M_1$ = molarity of $H_2SO_4$ solution = ?

$V_1$ = volume of $H_2SO_4$ solution = 50.0 ml

$n_2$ = acidity of $NaOH$ = 1

$M_1$ = molarity of $NaOH$ solution = 0.375 M

$V_1$ = volume of $NaOH$ solution = 62.5 ml

Putting in the values we get:

$2\times M_1\times 50.0=1\times 0.375\times 62.5$

$M_1=0.234M$

Therefore concentration of $H_2SO_4$ is 0.234 M

6 0

3 years ago

The building block of all matter is a) water b) oxygen c) an atom d) an element

jeka94

My best try would be d an element

4 0

3 years ago

2.50 liter of a gas has a pressure of 165. kPa at 25.0°C. If the pressure increases to 600. kPa and the temperature to 100.0°C,

tiny-mole [99]

Answer:

0.861 L

Explanation:

We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Convert the temperatures to degrees Kelvin.

25.0°C -> 298 K, 100.0°C -> 373 K

Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:

(165(2.5))/298 = (600(V₂))/373

V₂ = (165(2.5)(373))/(298(600))

V₂ = 0.861 L

7 0

3 years ago

A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes

ziro4ka [17]

Answer:

PCO2 = 0.6 25 atm

PSO2 = 1.2 75 atm

PO2 = 0.6 atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For 1 mole CS2 we need 3 moles O2 to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6 atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0

3 years ago

Ga3+ and Br1- is what formula?

Sladkaya [172]

$\text{GaBr}_3$

8 0

3 years ago