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3 years ago

A 25.00 mL sample of the ammonia solution

Chemistry

1 answer:

musickatia [10]3 years ago

3 0

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 = 250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

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Answer : (C) "Higher frequencies have larger spaces between lines".

Explanation:

In Young's experiment, the condition for constructive interference is given by :

$dsin\theta=n\lambda$ .........(1)

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We also know that, $c=\nu \lambda$

$\lambda=\dfrac{c}{\nu}$

where,

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So, equation (1) turns into

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So, correct option is (C).

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2 years ago

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REY [17]

Answer:

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Explanation:

Know the rules of multiplying wth perentheses.

- Hope that helped! Please let me know if you need further explanation.

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3 years ago

How much energy is required to raise the temperature of 10.6 grams of gaseous neon from

Alona [7]

Answer:

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Explanation:

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Let $m$ denote the mass of a sample of specific heat $C$ . Energy required to raise the temperature of this sample by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ .

For the neon gas in this question:

$c = 1.03\; \rm J \cdot g^{-1}\cdot K^{-1}$ .
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$\Delta T = (37.9 - 20.0)\; \rm K = 17.9\; \rm K$ .

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3 years ago

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strojnjashka [21]

Answer:

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