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DanielleElmas [232]
3 years ago
13

A student is given a stock solution of 1.00 M NaCl in water. They are asked to make 5 mL of 0.0500 M NaCl. How much of the stock

solution should they dilute to 5 mL to make the correct concentration?
a) 10.0 mL
b) 55.5 mL
c) 1.00 mL
d) 35.3 mL
e) 27.8 mL
Chemistry
1 answer:
Harrizon [31]3 years ago
3 0

Answer:

The answer to your question is None of your answers is correct, maybe the data are wrong.

Explanation:

Data

Concentration 1 = C1 = 1 M

Volume 2 = 5 ml

Concentration 2 = 0.05 M

Volume 1 = x

To solve this problem use the dilution formula

             Concentration 1 x Volume 1 = Concentration 2 x Volume 2

Solve for Volume 1

              Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1

Substitution

              Volume 1 = (0.05 x 5) / 1

Simplification

              Volume 1 = 0.25/1

Result

               Volume 1 = 0.25 ml

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Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g

Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.

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Answer:

The diameter of the hydrogen \mathbf{d =1.0605 \times 10^{-10}\ m}

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

L = \dfrac{nh}{2 \pi}

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L = mvr.

replacing the value of L into the previous equation, we have:

mvr= \dfrac{nh}{2 \pi}

v= \dfrac{nh}{2 \pi mr} ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}   ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2

ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}

r =\dfrac{n^2h^2}{4 \pi^2 mke^2}

For ground-state n = 1

h = (6.625 \times 10^{-34} \ J.s)^2

m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)

Ke = (1.6 \times 10^{-19} \ C)^2

r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}

r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}

\mathbf{r = 5.3025 \times 10^{-11} \ m}

Therefore, the diameter of hydrogen d = 2r

\mathbf{d = ( 2 \times  5.3025 \times 10^{-11} \ m})}

\mathbf{d =1.0605 \times 10^{-10}\ m}}

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3 years ago
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We are given the amount of ammonia formed from the reaction. This is where we start our calculations.

 

0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2

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