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A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m

natima [27]

Answer:

a. 0.4544 N

b. $5.112 \times 10^{-5 M}$

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

$H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O$

$NaOH\ Mass = Normality \times equivalent\ weight \times\ volume$

$= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL$

= 0.27264 g

$NaOH\ mass = \frac{mass}{molecular\ weight}$

$= \frac{0.27264\ g}{40g/mol}$

= 0.006816 mol

Now

Moles of $H_2SO_4$ needed is

$= \frac{0.006816}{2}$

= 0.003408 mol

$Mass\ of\ H_2SO_4 = moles \times molecular\ weight$

$= 0.003408\ mol \times 98g/mol$

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

$= \frac{acid\ mass}{equivalent\ weight \times volume}$

$= \frac{0.333984 g}{49 \times 0.015}$

= 0.4544 N

b. And, the acid solution molarity is

$= \frac{moles}{Volume}$

$= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}$

= 0.00005112

= $5.112 \times 10^{-5 M}$

We simply applied the above formulas

4 0

3 years ago

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst

expeople1 [14]

Answer:

s= 20.4 m

Explanation:

First lets write down equations for each ball:

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s

now we just have to put that time in any of those equations an get distance from the ground:

s = 30 + 5*2 -1/2*9.81 *2^2

s= 20.4 m

6 0

2 years ago

In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o

lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

= $10\times 101325 \ Pa$

= $1013250 \ Pa$

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

= $1 \ m^3$

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ $PV=nRT$

o,

⇒ $n=\frac{PV}{RT}$

By substituting the values, we get

$=\frac{1013250\times 1}{8.3145\times 298}$

$=408.94 \ moles$

As we know,

⇒ $Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}$

or,

⇒ $MW=\frac{m}{n}$

$=\frac{11.5}{408.94}$

$=0.02812 \ Kg/mol$

$=28.12 \ g/mol$

3 0

2 years ago

At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing

Inessa [10]

Answer:

vB = - 0.176 m/s (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)

⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)

⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒ vB = - 0.176 m/s (↓-)

The pic can show the question.

7 0

3 years ago

Read 2 more answers

A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =

oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0

2 years ago