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2 years ago

What is the mass of a sample of NH3 containing 6.3 × 1024 molecules of NH3?

Chemistry

1 answer:

quester [9]2 years ago

7 0

Answer:

178 grams

Explanation:

It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).

Using cross multiplication:

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 6.3 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(6.3 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 10.46 mol.

∴ The no. of grams of NH₃ present = no. of moles x molar mass = (10.46 mol)(17.0 g/mol) = 177.8 g ≅ 178 g.

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Is utensils a substance homogeneous mixture or heterogeneous mixture

Phoenix [80]

A pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.

7 0

2 years ago

Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber bra

Goryan [66]

Answer:

0.921 J/g degrees C

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation). Therefore, heat given off by the brakes = −heat taken in by tires, or:

−qbrakes=qtires

The equation used to calculate the quantity of heat energy exchanged in this process is:

−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires

First we must convert the mass of the tires and the brakes from kg to g.

massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g

masstires=123 kg×1,000. g1 kg=1.23×105 g

Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.

−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)

ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C

The answer should have three significant figures, so round to 0.921Jg∘C.

6 0

2 years ago

An ice cube measuring 5.80 cm by 5.80 cm by 5.80 cm has a density of 0.917 g/mL What is the mass?

OLga [1]

Answer: 178.9 g

Explanation:

Density = $\frac{mass}{volume in mL}$

find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³

1.0 cm³ = 1.0 mL

so 195.112 cm³ = 195.112 mL

plug value into density equation:

0.917 g/mL = (mass) / (195.112 mL)

and solve for mass!

3 0

2 years ago

In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OAT

dmitriy555 [2]

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy

$\Delta H$ = enthalpy change

$\Delta S$ = entropy change

T = temperature in Kelvin

When $\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

For the given reaction 1: $glucose+Pi\rightarrow glucose-6-phosphate+H_2O$ $\Delta G=+13.8kJ/mol$

As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

6 0

2 years ago

Compound a on ozonolysis yields acetophenone and propanal. what is the structure of compound a? 1-phenyl-1-hexene 1-phenyl-2-pen

astra-53 [7]

Answer:
2-Phenyl-2-Penetene on ozonolysis yields acetophenone and propanal.

Explanation:
The tricky way to solve such questions is to simply break the double bond in alkene and place oxygen atom at each broken half double bond making it carbonyl group. The reaction of given statement is as follow,

6 0

2 years ago