-What number is 66.3% of 56? -Find a number so that 25.5% of it is 10.

Help me please( ╹▽╹ )

xeze [42]

Lower temperature

Let's verify

Pressure=P
volume=V
Temperature=T

As per Boyles law

$\\ \rm\Rrightarrow V\propto \dfrac{1}{P}$

As per Charles law

$\\ \rm\Rrightarrow V\propto T$

$\\ \rm\Rrightarrow T\propto \dfrac{1}{P}$

So

At higher altitudes lower the pressure so lower the temperature

4 0

3 years ago

Which statement about gases is true? A. They are made up of particles that always move very slowly. B. They are made up of parti

rosijanka [135]

Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

A gas is a state of aggregation of matter in which, under certain conditions of temperature and pressure, its molecules interact weakly with each other, without forming molecular bonds, adopting the shape and volume of the container that contains them and tending to separate everything possible because of its high concentration of kinetic energy.

The molecules of a gas are practically free and have the ability to be distributed throughout the space in which they are contained because the gravitational forces and attraction between them are practically negligible compared to the speed at which they move. .

Therefore, gas molecules do not travel specific trajectories or vibrate in a stationary position, instead they move quickly and randomly through the entire space of the container that contains them.

7 0

3 years ago

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If you have a gas sample with an initial pressure of 16 atm, an initial volume of 20 L and a temperature of 500 K, what will the

DiKsa [7]

Answer:

T₂ = 218.75 K

Explanation:

Given data:

Initial volume = 20 L

Initial pressure = 16 atm

Initial temperature = 500 K

Final temperature = ?

Final volume = 35 L

Final pressure = 4 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

P₁V₁/T₁ = P₂V₂/T₂

T₂ = P₂V₂ T₁/ P₁V₁

T₂ = 4 atm × 35 L × 500 K / 16 atm × 20 L

T₂ = 70000 atm .L. K / 320 atm.L

T₂ = 218.75 K

5 0

3 years ago

What features do all the resonance forms of a molecule or ion have in common?

NeX [460]

Answer/Explanation:

Things that will vary include the distribution of electrons between atoms, lone pairs, and bonds, the number of single vs double or triple bonds, and the formal charges on atoms in the structure.

All resonance forms of the same molecule or ion must have the same number of atoms, connected in the same way, and the same number of total electrons.

5 0

3 years ago

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IV.2. The following problem considers the combustion of butane in a torch. Molecular weight of butane 58.0 g/mol. 2C4H10 + 1302

Anarel [89]

Answer:

(a) Oxygen

(b) 0.84 g

(c) 2.54 g

Explanation:

(a)

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For butane

Given mass = 1.00 g

Molar mass of butane = 58.0 g/mol

Moles of butane = 1.00 g / 58.0 g/mol = 0.0172 moles

Given: For $O_2$

Given mass = 3.00 g

Molar mass of $O_2$ = 32.0 g/mol

Moles of $O_2$ = 3.00 g / 32.0 g/mol = 0.09375 moles

According to the given reaction:

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

2 moles of butane react with 13 moles of $O_2$

1 mole of butane react with 13/2 moles of $O_2$

0.0172 moles of butane react with (13/2)*0.0172 moles of $O_2$

Moles of $O_2$ required = 0.1118 moles

Available moles of $CuSO_4$ = 0.09375 moles

Limiting reagent is the one which is present in small amount. Thus, $O_2$ is limiting reagent. (0.09375 < 0.1118 )

(b)

The formation of the product is governed by the limiting reagent. So,

13 moles of $O_2$ react with 2 moles of butane

1 mole of $O_2$ react with 2/13 moles of butane

0.09375 mole of $O_2$ react with (2/13)*0.09375 moles of butane

Moles of butane used = 0.0144 moles

Molar mass of butane = 58.0 g/mol

Mass of butane used = Moles × Molar mass = 0.0144 × 58.0 g = 0.84 g

(c)

13 moles of $O_2$ on reaction forms 8 moles of carbon dioxide

1 mole of $O_2$ on reaction forms 8/13 moles of carbon dioxide

0.09375 mole of $O_2$ on reaction forms (8/13)*0.09375 moles of carbon dioxide

Moles of carbon dioxide obtained = 0.05769 moles

Molar mass of $CO_2$ = 44.0 g/mol

Mass of $CO_2$ = Moles × Molar mass = 0.05769 × 44.0 g = 2.54 g

6 0

3 years ago