How many molecules are in 2.50 moles of methane?

Write the nuclear symbol of oxygen-16?

Nina [5.8K]

Answer:

Oxygen-16 (16o) is a stable isotope of oxygen, having 8 neutrons and 8 protons in its nucleus.

Explanation:

5 0

3 years ago

A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new

Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

$PV=nRT$

P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

$V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L$

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

$PV=nRT$

P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

$n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles$

Thus the now the container contains 2.07 moles.

6 0

3 years ago

Blank is how strong gravity pulls on an object

zheka24 [161]

It is either netforce or gforce

3 0

3 years ago

What happens to the ions when the drain is turned off? please help :] ❤️

kkurt [141]

Answer:

In a long channel MOSFET, the width of the pinch-off region is assumed small relative to the length of the channel. Thus, neither the length nor the voltage across the inversion layer change beyond the pinch-off, resulting in a drain current independent of drain bias. Consequently, the drain current saturates.

Explanation:

pls Mark my answer in brainlist and follow me

5 0

3 years ago

(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S

Sunny_sXe [5.5K]

These are two questions and two answers

Answer:

Question 1:

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)

Question 2:

0.201 M

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

sulfuric acid + potassium hydroxide → potassium sulfate + water

↑ ↑ ↑ ↑

acid base salt water

2) Skeleton equation (unbalanced)

H₂SO₄ + KOH → K₂SO₄ + H₂O

#) Balanced chemical equation (including phases)

H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

2) Moles of H₂SO₄:

V = 0.750 liter
M = 0.480 mol/liter
M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

3) Moles of KOH:

V = 0.700 liter
M = 0.290 mol/liter
M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

4) Determine the limiting reagent:

a) Stoichiometric ratio:

1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄ is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

Total volume = 0.700 liter + 0.750 liter = 1.450 liter

Concetration = M

M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

6 0

3 years ago