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3 years ago

PLEASE HELP ME ASAPPPPP How many moles are in 10.25 grams of NH3?

1 answer:

8 0

$10.25g NH3*( \frac{1mol NH3}{17.04g NH3} )=0.6015$

$0.6015 = 6.015*10^{-1} mol NH3$

$6.015*10^{-1} mol NH3$ is the answer.

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What is the equation for determining gas pressure?

Advocard [28]

Answer:

Explanation:

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volume in liters, 'n' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmospheres per moles Kelvin)

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3 years ago

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Which of the following terms best describes any solution?

baherus [9]

I believe D hopefully this helps

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4 years ago

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A 4.305-g sample of a nonelectrolyte is dissolved in 105 g of water. the solution freezes at -1.23c. calculate the molar mass of

amid [387]

The answer is 62.00 g/mol.
Solution:
Knowing that the freezing point of water is 0°C, temperature change Δt is
Δt = 0C - (-1.23°C) = 1.23°C
Since the van 't Hoff factor i is essentially 1 for non-electrolytes dissolved in water, we calculate for the number of moles x of the compound dissolved from the equation
Δt = i Kf m
1.23°C = (1) (1.86°C kg mol-1) (x / 0.105 kg)
x = 0.069435 mol
Therefore, the molar mass of the solute is
molar mass = 4.305g / 0.069435mol = 62.00 g/mol

6 0

3 years ago

1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm

sashaice [31]

1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm

Ideal gas equation: pV = nRT => n = pV / RT

R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm

=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol

2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type

X (+) + O2 (g) ---> X2O or

2 X(2+) + O2(g) ----> X2O2 = 2XO or

4X(3+) + 3O2(g) ---> 2X2O3

In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)

So, lets probe those 3 cases.

3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X

Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol

That does not correspond to any of the metal with valence 1+

So, now probe the case 2.

4) Case 2:

2moles X metal / 1 mol O2(g) = x / 0.01226 mol

=> x = 2 * 0.01226 = 0.02452 mol

And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol

That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.

4) Case 3

4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635

atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol

That does not correspond to any metal.

Conclusion: the identity of the metallic element could be titanium.

5 0

3 years ago

If I use the dropper to add 10 mL of drink mix to 50 mL of water, what is the

Yuliya22 [10]

Answer:

16.7%

Explanation:

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

Volume of drink mix = 10 mL

Volume of water = 50 mL

Percentage of drink mix =?

Next, we shall determine the total volume of the solution. This can be obtained as follow:

Volume of drink mix = 10 mL

Volume of water = 50 mL

Total volume of solution =?

Total volume of solution = (Volume of drink mix) + (Volume of water)

Total volume of solution = 10 + 50

Total volume of solution = 60 mL

Finally, we shall determine the percentage of the drink mix in the solution. This can be obtained as follow:

Volume of drink mix = 10 mL

Total volume of solution = 60 mL

Percentage of drink mix =?

Percentage of drink mix = Vol. of drink mix / total volume × 100

Percentage of drink mix = 10/60 × 100

Percentage of drink mix = 16.7%

4 0

3 years ago