What is the molarity of 2 moles of iron oxide,FeO in 8 Liters of water

a sample of the hydrate of sodium carbonate has a mass of 8.85 g. it loses 1.28 g when heated. find the molar ratio of this comp

xenn [34]

Answer:

Molar ratio of the compound is 1:1 and the type of hydrate is Mono hydrate.

Explanation:

From the given,

Mass of sodium carbonate $Na_{2}CO_{3}.XH_{2}O$ = 8.85 g

Loss mass $H_{2}O$ = 1.28 g

Actual weight of sodium carbonate = 8.85 g - 1.28 g = 7.57 g

$7.57 g Na_{2}CO_{3} \times \frac{1mol}{106 g} =\frac{0.0714}{0.0714}=1$

$1.28g H_{2}O \times \frac{1mol}{18 g} =\frac{0.0711}{0.0714}=1$

Therefore, the compound has only one water molecule.

Molecular formula of the compound is $Na_{2}CO_{3}.H_{2}O$ an name of the compound is sodium carbonate mono hydrate.

Hence, the type of the compound is Mono hydrate.

8 0

3 years ago

Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

3 years ago

Which processes of the water cycle involve the movement of water in the liquid state?

dangina [55]

Answer:

Surface runoff and condensation

Explanation:

Let's define each of the given processes in order to understand them better:

evaporation is a process in which liquid phase transforms into a gas phase;
precipitation is a process in which we produce a solid phase, usually this is the case when we precipitate a salt out of a solution, analogy of precipitation for water would be transformation from a liquid to a solid phase, such as freezing;
surface runoff is a process in which water flows over the surface of a land without any change in its phase;
condensation is a process in which a gas transforms into a liquid.

All in all, notice that surface runoff keeps water in its liquid state, while all the other three options consider phase change. The only phase change of interest is condensation: we produce liquid water from water vapor and then we can analyze its movement in the liquid state.

6 0

3 years ago

Hydrogen bonds are

ella [17]

Answer:

c

Explanation:

hydrogen bonds are initially weak and collectively strong because they are partially positive ions

3 0

3 years ago

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C

laila [671]

Answer:

First choice: 99 J/K

Explanation:

1) First law of thermodynamic (energy balance)

Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

2) Energy change of each substance:

General formula:

Heat released or absorbed = mass × Specific heat × change in temperature

density of water: you may take 0.997 g/ ml as an average density for the water.

mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

Specif heat of water: 1 cal / g°C

Heat released by the hot water:

Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

Heat absorbed by the cold water:

Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

Heat absorbed by the calorimeter

Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

4) Balance

Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

Convert to cal / K to Joule / K

1 cal = 4.18 Joule

23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0

4 years ago