Is LiOH soluble or insoluble?

Pls help i will mark you the brainlest

kap26 [50]

Answer:

Maintain constant velocity

8 0

3 years ago

Give an example of competition in an ecosystem

neonofarm [45]

hbzjaj

Explanation:

competition for food

5 0

3 years ago

When you mixed 20 grams of magnesium and an excess of nitric acid, 1.7 grams of hydrogen was actually produced. What is the perc

Kitty [74]

Based on the balanced chemical reaction presented above, every mole of magnesium (Mg) yields one mole of diatomic hydrogen (H2). When converted to masses, every 24.3 grams of magnesium yields 2 grams of hydrogen.

From the given, there are 20 grams of magnesium available for the reaction. With this amount, the expected yield of hydrogen is 1.646 grams. To calculate the percent yield, divide the actual yield to the hypothetical yield.

*The case is impossible because the actual yield is greater than the theoretical yield.

If we assume that there had been a typographical error and that the actual yield is 0.7 grams instead of 1.7 grams, the percent yield becomes 42.5%. Thus, the answer is letter E.

6 0

3 years ago

Read 2 more answers

What evidence did Wegener use to support his continental drift<br> hypothesis?

Mrac [35]

Answer: Coastlines.

Explanation: South America's and Africa's shoreline interlocks perfectly. Which implies they must have been joined together in history.

6 0

2 years ago

Reaction of 0.028 g of magnesium with excess hydrochloric acid generated 31.0 mL of hydrogen gas. The gas was collected by water

MA_775_DIABLO [31]

Explanation:

(a) It is given that magnesium is reacted with hydrochloric acid and the hydrogen evolved is collected at top. This means that hydrochloric acid will be present in a solution (HCl + Water) and the solvent will be water.

Due to evaporation some amount of water will have evaporated and would be present in vapor phase. Therefore, when the reaction occurs only hydrogen will not be present in vapor phase but, will be accompanied by water vapors as well .

Hence, Dalton's law the total pressure of the system will be sum of pressure exerted by hydrogen gas and pressure exerted by water vapors .

Let us assume that the partial pressure of hydrogen gas be " $P_H_{2}$ "

And, the partial pressure of water will be nothing but the vapor pressure of water,

Vapor pressure of water = $P_{water}$

= 19.8 mm Hg

Total pressure of the system = 746 mm Hg

Total pressure = $P_H_{2} + P_{water}$

746 = $P_H_{2}$ + 19.8

or, $P_H_{2}$ = 746-19.8

= 726.2 mm Hg

Hence, partial pressure of hydrogen gas is 726.2 mm Hg.

(b) To calculate volume at STP, we will first calculate at $22^{o}C$ and 726.2 mm Hg and than convert it to STP conditions.

Therefore, to calculate volume at $22^{o}C$ and 726.2 mm Hg we will make use of ideal gas law as follows.

P = 726.2 mm Hg

= $\frac{726.2}{760}$

= 0.955 atm

T = $22^{o}C$

= 22+273.15 = 295.15 K

V = 31 ml = $31 \times 10^{-3}$ Litre

According to the ideal gas law ,

PV = nRT

where, P = pressure of the system ,

V = volume of the gas

N = number of moles

R = 0.0821 liter atm/mole K

T = Temperature

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

$0.955 \times 31 \times 10^{-3} = N \times 0.0821 \times 295.15$

N = $1.222 \times 10^{-3}$ moles

Now, the moles of hydrogen won't change. Therefore, let us calculate volume at STP of $1.222 \times 10^{-3}$ moles of hydrogen.

Now, at STP T = 273.15 K , P = 1 atm and N = $1.222 \times 10^{-3}$ moles

$1 \times V = 1.222 \times 10^{-3} \times 0.0821 \times 273.15 K$

V = 0.027398 Litre

= $0.027398 \times 1000$ (as 1 L = 1000 ml)

= 27.398 ml

Therefore, volume of hydrogen at STP is 27.398 ml .

(c) Now, we can write the the reaction for this case as follows.

$Mg + 2HCl \rightarrow MgCl_{2} + H_{2}$

As, weight of magnesium = 0.028 grams

Molar mass of magnesium = 24.3 grams/mole

Number of moles of magnesium = $\frac{mass}{\text{molar mass}}$

= $\frac{0.028}{24.3}$

= $1.15226 \times 10^{-3}$ moles

Since, it can be seen from the reaction that

1 mole of Magnesium = 1 mole of hydrogen

and, moles of hydrogen = $1.15226 \times 10^{-3}$ moles

= 0.001523 moles

Hence, theoretical number of moles of hydrogen that can be produced from 0.028 grams of Mg is 0.001523 moles

8 0

3 years ago