How many moles are in 60.66 g CF2Cl2

If a object started moving have did it’s velocity change?

Otrada [13]

Answer:

the velocity did change

Explanation:

velocity is the rate of change of its position with respect to a frame of reference.

3 0

4 years ago

2al(s)+3cl2(g)→2alcl3(s) δh∘ = -1408.4 kj. how much heat (in kilojoules) is released on reaction of 4.70 g of al?

Annette [7]

Answer:
- 122.58 kJ

Solution:
According to following equation,

2 Al + 3 Cl₂ → 2 AlCl₃ δH° = -1408.4 kJ

When,
54 g (2 mole) Al on reaction releases = 1408.4 kJ heat
So,
4.70 g of Al will release = X kJ of Heat

Solving for X,
X = (4.70 g × 1408.4 kJ) ÷ 54 g

X = - 122.58 kJ

7 0

4 years ago

The diagram below is an artist’s impression of a single atom of element Be. The neutrons are shown with stripes, the protons are

Ahat [919]

Answer is: beryllium-10.

The diagram shows that atom has 4 protons, 5 neutrons and 2 valence electrons.

Atomic number is the number of protons, which is characteristic of a chemical element, beryllium (Be) is an element with atomic number 4.

Two valence electrons means that atom is from 2. group of periodic table, only beryllium is from that group; sodium (1. group), boron (13. group) and carbon (14. group).

Beryllium-10 has 6 neutrons, so it is isotope (different number of neutrons or mass number).

3 0

3 years ago

Read 2 more answers

Which formula equation represents the burning of sulfur to produce sulfur dioxide?

Leni [432]

Answer:

S(s) + O2(g) --> SO2(g)

Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

Explanation:

The reaction is given as;

Sulfur + oxygen --> Sulphur dioxide

Sulphur = S

Oxygen = O2

Sulfur dioxide = SO2

So we have;

S(s) + O2(g) --> SO2(g)

The crrect option is option A. Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

5 0

3 years ago

Read 2 more answers

A sample of methane gas having a volume of 2.80 L at 25 degree C and 1.65 atm was mixed with a sample of oxygen gas having a vol

Svetradugi [14.3K]

Answer:

Volume of carbon dioxide = 2.47 L

Explanation:

Given that:

For methane gas:-

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25.0 + 273.15) K = 298.15 K

V = 2.80 L

Pressure = 1.65 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

1.65 atm × 2.80 L = n ×0.0821 L atm/ K mol × 298.15 K

⇒n for methane gas = 0.1887 mol

For oxygen gas:-

Temperature = 31 °C

T = (31 + 273.15) K = 304.15 K

V = 35.0 L

Pressure = 1.25 atm

Using ideal gas equation as:

$PV=nRT$

Applying the equation as:

1.25 atm × 35.0 L = n ×0.0821 L atm/ K mol × 304.15 K

⇒n for oxygen gas = 1.7521 mol

According to the reaction:-

$CH_4+2O_2\rightarrow CO_2+2H_2O$

1 mole of methane gas reacts with 2 moles of oxygen gas

Also,

0.1887 mole of methane gas reacts with 2*0.1887 moles of oxygen gas

Moles of oxygen gas = 0.3774 moles

Available moles of oxygen gas = 1.7521 moles

Limiting reagent is the one which is present in small amount. Thus, methane gas is limiting reagent.

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

1 mole of methane gas on reaction forms 1 mole of carbon dioxide

Thus,

0.1887 mole of methane gas on reaction forms 0.1887 mole of carbon dioxide

Moles of carbon dioxide = 0.1887 moles

Given that:- Pressure = 2.50 atm

Temperature = 125 °C

T = (125 + 273.15) K = 398.15 K

Using ideal gas equation as:

$PV=nRT$

Applying the equation as:

2.50 atm × V = 0.1887 moles ×0.0821 L atm/ K mol × 398.15 K

Volume of carbon dioxide = 2.47 L

7 0

3 years ago