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2 years ago

After 7.9 grams of sodium are dropped into a bathtub full of water how many grams of hydrogen gas are released

Chemistry

1 answer:

Nataliya [291]2 years ago

6 0

Answer:

The grams of hydrogen gas that are released is 0.343 gram

Explanation:

Given that:

2 Na + 2h₂0 ⇒ 2Na0H +N₂

Molecules of Na = 7.9/2.3 = 0.343 mol Na

Now,

2 molecules of Na produces one mole h₂

0.343 mol of Na produce 0.343/2 = 0.1717 mol h₂

h2 = molecules * mw (molecular weight) =0.1717 * 2 =0.343 grams

Therefore the grams of hydrogen gas released is = 0.343 grams

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A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.7 nm.93.7 nm. What is the final state of the hy

sammy [17]

Answer:

Explanation:

Given that the formula is;

1/λ= R(1/nf^2 - 1/ni^2)

λ = 93.7 nm or 93.7 * 10^-9 m

R= 1.097 * 10^7 m-1

nf = ?

ni = 1

From;

ΔE = hc/λ

ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9

ΔE = 21 * 10^-19 J

ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)

21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)

-0.963 = (1/nf^2 - 1)

-0.963 + 1 = 1/nf^2

0.037 = 1/nf^2

nf^2 = (0.037)^-1

nf^2 = 27

nf = 5

7 0

2 years ago

Using the solubility curve, what is the effect of increased temperature on the solubility of KBr in 100 grams of water? The solu

galina1969 [7]

The graph is needed to answer this question.

Solubility may increase or decrease with temperature depending on the properties of the solute and the solvent.

It is quite common that the solubility of the ionic compounds, like KBr, in water increases with temperature.

Use your solubility curve for the KBr and you wiil see a line that starts at a solubility a little greater than 50 grams of the salt in 100 grams of water for temperaute 0°C and increase linearly until almost 100 grams of the salt in 100 grams of water at 100°C.

So, in this case you can affirm that the solubility of KBr increases with the temperature.

Answer: the second option: the solubility increases.

7 0

2 years ago

Read 2 more answers

Question 3 The following chart shows where the various forms of carbon are found. What correction needs to be made? Atmosphere B

bagirrra123 [75]

Answer:

i think it is A

Explanation:

hope this helps

6 0

2 years ago

Read 2 more answers

When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po

SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide ( $C_{7}H_{7}NO$ ) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is $2.7^{o}C$ lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride ( $NH_{4}Cl$ ) are dissolved in the same mass of X, the freezing point of the solution is $9.9^{o}C$ lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

Moles of benzamide = $\frac{mass}{\text{Molar mass of benzamide}}$

= $\frac{70.4 g}{121.14 g/mol}$

= 0.58 mol

Now, we will calculate the molality as follows.

Molality = $\frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}$

= $\frac{0.58 mol}{0.85 kg}$

= 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

dT = $i \times K_{f} \times m$ ,

where, dT = change in freezing point = $2.7^{o}C$

i = van't Hoff factor = 1 for non dissociable solutes

$K_{f}$ = freezing point constant of solvent

m = 0.6837

Therefore, putting the given values into the above formula as follows.

dT = $i \times K_{f} \times m$ ,

$2.7^{o}C = 1 \times K_{f} \times 0.6837 m$

$K_{f}$ = 3.949 C/m

Now, we use this $K_{f}$ value for calculating i for $NH_{4}Cl$

So, moles of ammonium chloride are calculated as follows.

Moles of $NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}$

= 1.316 mol

Hence, calculate the molality as follows.

Molality = $\frac{1.316 mol}{0.85 kg}$

= 1.5484

It is given that value of change in temperature (dT) = $9.9^{o}C$ . Thus, calculate the value of Van't Hoff factor as follows.

dT = $i \times K_{f} \times m$

$9.9^{o}C = i \times 3.949 C/m \times 1.5484 m$

i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0

2 years ago

Atmospheric pressure decreases as altitude increases. in other words, there is more air pushing down on you at sea level, and th

LuckyWell [14K]

Boiling point is the temperature at which vapour pressure equals atmospheric pressure.

As, we move at higher altitudes, atmospheric pressure decreases. Hence, temperature to reach the boiling point will decrease.

Further, boiling point is higher for longer chain compounds. Hence, octane (C8H18) and octanol (C8H17OH) will have higher boiling point as compared to hexane (C6H14). Further, intermolecular forces of interaction are more stronger in octanol, due to presence of OH group, as compared to octane.

Hence, boiling points will be in following order:
Octanol > Octane > Hexane

Thus, hexane will boil first, followed by octane and lastly octanol.

7 0

2 years ago