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maks197457 [2]
3 years ago
13

Day 1 code word for science?

Chemistry
1 answer:
Flura [38]3 years ago
8 0

Answer:

hail Hydra kandksmak Yes

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What is the mass of 925.4 L of hydrogen gas at STP?
Ahat [919]

Answer:

If this is an idea gas then 1mol takes up 22.4L.

So, knowing how many L you have you can figure out how many  mole syou have by doing a simple equation:

\frac{1mol}{y} =\frac{22.4L}{925.4L}

Solve for y.

Then, since you know how many moles you have use the ptable https://ptable.com/#Properties to figure out the mass in grams.

NOTE: The ptable tells you that 1mol of H = 1g.....so this should be an easy calculation :) enjoy

8 0
3 years ago
How many sulfur atoms are in 0.45 mol BaSO4??
Lemur [1.5K]

(0.45 mol BaSO_{4})*(\frac{6.022*10^23 molecules}{mol})*(\frac{1 atom S}{1 molecule BaSO_{4}}) = 2.7*10^{23}

4 0
3 years ago
Read 2 more answers
Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz
stiv31 [10]

Answer: Freezing point of a solution will be -1.16^0C

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}

(5.50-T_f)^0C=6.68

T_f=-1.16^0C

Thus the freezing point of a solution will be -1.16^0C

3 0
3 years ago
The volume of a cube is calculated by using the following formula: l x h x w. The length (l), height (h), and width (w) of a met
aliya0001 [1]

Answer:

8.909466637830

Explanation:

herp derp

8 0
3 years ago
A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution
Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

<em></em>

I hope it helps!

7 0
3 years ago
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