Describe the shape of the graph during the period when water is freezing. Why the graph is shaped this way?

Explain why you can not use a physical reaction to separate elements that are chemically bonded together

Rainbow [258]

Their atoms are too tightly bonded to each other for them to be separated by a physical reaction

3 0

3 years ago

Calculate the mass of 120cc of nitrogen at stp .how many numbers of molecules are prrsent

Ratling [72]

Answer:

3.2 x 10²¹molecules

Explanation:

Given parameters:

Volume of nitrogen gas = 120cm³

Unknown:

Mass of nitrogen gas = ?

Number of molecules = ?

Solution:

To solve this problem, note that;

1 mole of gas occupies a volume of 22.4L at STP

Now;

convert 120cm³ to L ;

1000cm³ = 1L

120cm³ gives 0.12L

Since;

22.4L of gas has 1 mole at STP

0.12L of gas will have $\frac{0.12}{22.4}$ = 0.0054mole at STP

So;

Mass of N₂ = number of moles x molar mass

Molar mass of N₂ = 2(14) = 28g/mol

Mass of N₂ = 0.0054 x 28 = 0.15g

Now;

1 mole of a gas will have 6.02 x 10²³ molecules

0.0054 mole of N₂ will contain 0.0054 x 6.02 x 10²³ =

3.2 x 10²¹molecules

8 0

3 years ago

What does E+mc^2 mean and how it came to be

Shalnov [3]

E = mc^2<span> is an equation derived by the twentieth-century physicist Albert Einstein, in which E represents units of energy, m represents units of mass, and c 2 is the speed of light squared, or multiplied by itself.</span>

4 0

4 years ago

For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous pro

Maslowich

Answer and Explanation:

a. Solid melting below its melting point ⇒ Nonspontaneous

The process is spontaneous <u>above</u> the melting point.

b. Gas condensing below its condensation point ⇒ Spontaneous

Below the condensation point, the subtance is liquid (it condenses)

c. Liquid vaporizing above its boiling point ⇒ Spontaneous

A liquid vaporizes at a temperature above the boiling point, it passes from liquido to the gas state

d. Liquid freezing below its freezing point ⇒ Spontaneous

A liquid freezes at a temperature below its freezeing point, it passes from the liquid to the solid state.

e. Liquid freezing above its freezing point ⇒ Nonspontaneous

The freezing is spontaneous below the freezing point.

f. Solid melting above its melting point ⇒ Spontaneous

A solid melts at a temperature above the melting point

g. Liquid and gas together at boiling point with no net condensation or vaporization ⇒ Equilibrium system

The systems is at equilibrium: there is no net change toward the liquid or towards the gas state.

h. Gas condensing above its condensation point ⇒ Nonspontaneous

The condensation is spontaneous at a temperature below the condensation point.

i. Solid and liquid together at the melting point with no net freezing or melting ⇒ Equilibrium system

The system is at equilibrium: there is no net change towards the solid or the liquid state.

5 0

3 years ago

If the crucible originally weighs 3.715 g and 2. 000 g of hydrate are added to it , what is the weight of the water that is lost

dimulka [17.4K]

For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.

From the problem, you know that 1) = 2.000 g.

Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).

1) = 5.022 g - 3.715 g = 1.307 g

Now you can complete our original task.

Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.

4 0

4 years ago