Pleaaaaase hellllp asaaaap

Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr

Mrrafil [7]

Answer:

The value of an integer x in the hydrate is 10.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Molarity of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

$0.0366 M=\frac{n}{5.00 L}$

$n=0.0366 M\times 5 mol=0.183 mol$

Mass of hydrated sodium carbonate = n= 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

$n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}$

$0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}$

$106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}$

Solving for x, we get:

x = 9.95 ≈ 10

The value of an integer x in the hydrate is 10.

6 0

3 years ago

Which of the following is a property of a metal?

BaLLatris [955]

The correct answer is the 3rd option Ductility is a property of a metal. It is the ability of a material to deform under tensile stress.It is characterized by the ability to be stretched into wire-like form which is an ability of metals.

8 0

3 years ago

Which phrase describes a plateau?

Rina8888 [55]

Answer:

I think it's surface cut by streams

8 0

3 years ago

Read 2 more answers

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m

(b) $n_1$ =4 to $n_2$ = 20

$\frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2} )$ = 6580.62

Therefore, $\lambda$ = 1 / 6580.62 = 0.000152 m

(c) $n_1$ =3 to $n_2$ = 10

$\frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2} )$ = 11089.56

Therefore, $\lambda$ = 1 / 11089.56 = 0.00009 m

(d) $n_1$ =2 to $n_2$ = 1

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2} )$ = - 82257.75

Therefore, $\lambda$ = 1 /82257.75 = - 0.0000121 m

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0

1 year ago

A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro

Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

$NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)$

So, $OH^{-}$ is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of $OH^{-}$ increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of $OH^{-}$ and keep same equilibrium constant.

Therefore excess amount of $OH^{-}$ combines with $NH_{4}^{+}$ to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0

3 years ago