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3 years ago

How many mL of a 2.0M NaBr solution are needed to make 200.0 mL of 0.50M NaBr?

2 answers:

8 0

This is a classical C1V1 = C2V2 question. We are given the initial concentration (2.0M), the final concentration (0.5M) and the final volume (200mL). We are asked how many mL of 2.0 M NaBr we require to make up the final concentration and volume, so we are calculating the initial volume (V1)

Therefore:
C1V1 = C2V2
2.0M(V1) = (0.5M)(200.0 mL)
V1 = 50 mL
So 50 mL of NaBr is needed to make a dilution to 0.5M in 200 mL.

ololo11 [35]3 years ago

6 0

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

2.0 M x V1 = 0.50 M x 200 mL

V1 = 50 mL of <span>2.0M NaBr is needed</span>

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How does a conjugate acid differ from its conjugate base?

inna [77]

The conjugate acid is differ from its conjugate base as, conjugate acid is formed by strong base whereas conjugate base is formed by strong acid.

conjugate base is differ from conjugate acids by the presence of the proton. The conjugate acid is formed when proton is added to the bases whereas conjugate bases is formed when proton is released by the acids.

Example of corrugate acids are given below.

$NH_{3}$ → $NH_{2} ^{-} +H^{+}$

In the above example $NH_{2} ^{-}$ is conjugate acids.

Learn about conjugate acid

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2 years ago

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

When driving a truck,is fuel crucial?

Natali [406]

Answer:

The answer is yes.

Explanation:

When driving any type of motorized vehicle, fuel is importatn. Fuel is what it runs on, so without it it wouldn't run. With trucks a lot more is needed because trucks are bigger, and tend to carry more than a little car does.

6 0

4 years ago

What is the oxidation number of chlorine in HClO4?

Elza [17]

CIO4-=-1

CI=4O=-1

O has a 2- oxidation change so

CI+4(-2)=-1

CI+(-8)=-1

CI=-1+8=7

So the oxidation number of chlorine is 7 in this case

8 0

3 years ago

When a 120 g sample of aluminum absorbs 9612 of heat energy, its temperature increases from 25°C to 115°C. Find the specific hea

lesantik [10]

<h3>Answer:</h3>

0.89 J/g°C

<h3>Explanation:</h3>

Concept tested: Quantity of heat

We are given;

Mass of the aluminium sample is 120 g
Quantity of heat absorbed by aluminium sample is 9612 g
Change in temperature, ΔT = 115°C - 25°C

= 90°C

We are required to calculate the specific heat capacity;

We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.

That is;

Q = m × c × ΔT

Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.

Specific heat capacity, c = Q ÷ mΔT

= 9612 J ÷ (120 g × 90°C)

= 0.89 J/g°C

Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C

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3 years ago