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zheka24 [161]
2 years ago
10

Balence the equation : _____ Ga2O3 + _____ Li -----------> ______ Li2O + ______ Ga

Chemistry
2 answers:
Sergeeva-Olga [200]2 years ago
5 0

Answer:

1 Ga2O3 + 6Li——-> 3 Li2O + 2 Ga

Explanation:

NeX [460]2 years ago
3 0

Answer:

Ga2O3 + 6Li ------> 3Li2O + 2Ga

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A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan
Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = \frac{18.0 g}{60 g/mol}=0.3 mol

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}

Molarity of the urea solution ;

M=\frac{0.3 mol}{0.200 L}=1.50 M

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

m=d\times V=0.95 g/mL\times 200.0 mL=190 g

m = 190 g = 190 \times 0.001 kg = 0.19 kg

(1 g = 0.001 kg)

Molality of the urea solution ;

Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}

m=\frac{0.3 mol}{0.19 kg}=1.58 m

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

7 0
2 years ago
Plesse help asap!!!​
Naily [24]

Answer:

4.4×10² cm³

Explanation:

From the question given above, the following data were obtained:

Diameter (d) = 68.3 mm

Height (h) = 0.120 m

Volume (V) =?

Next, we shall convert the diameter (i.e 68.3 mm) to cm.

This can be obtained as follow:

10 mm = 1 cm

Therefore

68.3 mm = 68.3 mm / 10 mm × 1 cm

68.3 mm = 6.83 cm

Therefore, the diameter 68.3 mm is equivalent 6.83 cm.

Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:

1 m = 100 cm

Therefore,

0.120 m = 0.120 m/ 1 m × 100 cm

0.120 m = 12 cm

Therefore, the height 0.120 m is equivalent 12 cm.

Next, we shall determine the radius of the cylinder. This can be obtained as follow:

Radius (r) is simply half of a diameter i.e

Radius (r) = Diameter (d) /2

r = d/2

Diameter (d) = 6.83 cm

Radius (r) =?

r = d/2

r = 6.83/2

r = 3.415 cm

Finally, we shall determine the volume of the cylinder as follow:

Radius (r) = 3.415 cm

Height (h) = 12 cm

Volume (V) =?

Pi (π) = 3.14

V = πr²h

V = 3.14 × (3.415) ² × 12

V = 440 cm³

V = 4.4×10² cm³

Therefore, the volume of the cylinder is 4.4×10² cm³

8 0
2 years ago
Draw the lewis structure for bef2 in the box at the right, including lone pairs.
Kamila [148]

Answer :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and the unpaired electrons present in the molecule.

The given molecule is, BeF_2

Beryllium has '2' valence electrons and fluorine has '7' valence electrons. Beryllium is the central atom and fluorine is the terminal atom.

Total number valence electrons in BeF_2 = 2 + 2(7) = 16

According to the Lewis-dot structure, there are '4' number of bonding electrons and 12 number of non-bonding electrons (lone-pair).

The Lewis-dot structure is shown below.

3 0
2 years ago
1) Define water gas<br>2) The skuLL is divided into two parts name them​
Triss [41]

• a fuel gas consisting mainly of carbon monoxide and hydrogen, made by passing steam over incandescent coke

• the cranium and the mandible.

hope it helps...!!!

5 0
2 years ago
At 700 K, the reaction below has an Kp value of 54. An equilibrium mixture at this temperature was found to contain 0.933 atm of
kati45 [8]

Answer:

See explanation below

Explanation:

In this case, we have the equilibrium reaction which is:

H₂ + I₂ <------> 2HI       Kp = 54

Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:

Kp = PpHI² / PpH₂ * PpI₂

Solving for the partial pressure of iodine:

PpI₂ = PpHI² / PpH₂ * Kp

Replacing the given values, we have:

PpI₂ = (2.1)² / 0.933 * 54

PpI₂ = 4.41 / 50.382

PpI₂ = 0.088 atm

4 0
3 years ago
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