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3 years ago

. How many moles of ammonia gas, NH3, are required to fill a volume of 50 liters at STP?

Chemistry

2 answers:

murzikaleks [220]3 years ago

6 0

In STP every 22.4 litters is 1 mol

KATRIN_1 [288]3 years ago

5 0

Answer : The number of moles of ammonia gas is, 2.232 moles

Explanation : Given,

Volume of ammonia gas = 50 L

As we know that at STP,

1 mole of gas contains 22.4 L volume of gas.

As, 22.4 L volume of ammonia gas present in 1 mole of ammonia gas

So, 50 L volume of ammonia gas present in $\frac{50}{22.4}=2.232$ mole of ammonia gas

Therefore, the number of moles of ammonia gas is, 2.232 moles

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Which atom or ion is the largest? A. K B. K+ C. Ca D. Ca2+ E. Li

pishuonlain [190]

Answer:

A. K

Step-by-step explanation:

Remember the trends in the Periodic Table:

Atomic radii decrease from left to right across a Period.
Atomic radii increase from top to bottom in a Group.
Ionic radii of metal cations are smaller than those of their atoms.

Thus, the largest atoms are in the lower left corner of the Periodic Table.

The diagram below shows that K is closest to the lower left, so it is the largest atom. It is also larger than any of the cations.

5 0

3 years ago

What is the molarity (molar concentration, unit = M) of K+ found in 200 mL 0.2 M K2HPO4 solution?

enyata [817]

Answer:

0.4 M

Explanation:

The process that takes place in an aqueous K₂HPO₄ solution is:

K₂HPO₄ → 2K⁺ + HPO₄⁻²

First we calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution:

200 mL * 0.2 M = 40 mmol K₂HPO₄

Then we convert K₂HPO₄ moles into K⁺ moles, using the stoichiometric coefficients of the reaction above:

40 mmol K₂HPO₄ * $\frac{2mmolK^+}{1mmolK_2HPO_4}$ = 80 mmol K⁺

Finally we divide the number of K⁺ moles by the volume, to calculate the molarity:

80 mmol K⁺ / 200 mL = 0.4 M

5 0

2 years ago

The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m

MAXImum [283]

Answer : The molecular weight of this compound is 891.10 g/mol

Explanation : Given,

Mass of compound = 12.70 g

Mass of ethanol = 216.5 g

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure ethanol = $-117.300^oC$

$T_f$ = temperature of solution = $-117.431^oC$

$K_f$ = freezing point constant of ethanol = $1.99^oC/m$

i = van't hoff factor = 1 (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

$(-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}$

$\text{Molar mass of compound}=891.10g/mol$

Therefore, the molecular weight of this compound is 891.10 g/mol

7 0

2 years ago

A battery produces ____current, which is current that flows in only one direction.

fenix001 [56]

Answer:

The given blank can be filled with direct.

Explanation:

A battery refers to a chemical apparatus that is used to store electrical energy. A battery generates a direct current that is a current, which flows in only one direction. DC means direct current and it should exhibit fixed terminals, that is, negative and positive terminals, which cannot be altered.

As the batteries have fixed content, they possess fixed terminals, thus, they can store only direct current. Thus, the batteries produce direct current and produce a voltage in only one direction, and they exhibit a polarity of negative and positive.

5 0

2 years ago

What is the difference between intrusive igneous rocks and extrusive igneous rocks? We just learned this and I still don’t know

Reil [10]

The major visible difference between the two are crystal size, intrusive rocks have a larger crystal/grain texture due to the slow cooling of magma below the earth surface which encourages the growth of larger crystals, while extrusive rocks, because of the rapid cooling at/above the earth's surface does the opposite. Hope I helped

7 0

3 years ago

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