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3 years ago

If you had a glacier and an ice cube, how would the melting points of both differ?

Chemistry

1 answer:

Dima020 [189]3 years ago

8 0

Answer:

Due to how big the glacier is you would need more heat to melt it and since the ice cube is smaller a small amount of heat would melt it

Explanation:

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A Solution is a mixture of two or more substances_____

Ipatiy [6.2K]

Answer:

In the same phase

Explanation:

Because when you make a solution you have to keep them in the same phase or you will oof.

4 0

3 years ago

I have 345 mL of a 1.5 M NaCl solution. If I boil the water until the volume of the solution is 250 mL, what will the molarity o

likoan [24]

We know that molarity = mol/L, so:

1.5 M = x mol/0.345 L

1.5 M * 0.345 L = x mol => 0.5175 mol

0.5175 mol/0.250 L = 2.07 M

Your new molarity of the solution will be 2.07 M.

6 0

3 years ago

3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.

tiny-mole [99]

Answer:

The Avogadro's number is $N_A = 6.02289 *10^{23}$

Explanation:

From the question we are told that

The edge length is $L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}$

The density of the metal is $\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3$

The molar mass of Ba is $Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol$

Generally the volume of a unit cell is

$V = L^3$

substituting value

$V = [5.02 *10^{-10}]^3$

$V = 1.265*10^{-28}\ m^3$

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),

The volume of barium atom is

$V_a = \frac{V}{2} * 0.68$

substituting value

$V_a = \frac{ 1.265*10^{-28}}{2} * 0.68$

$V_a = 4.301 *10^{-29} \ m^3$

The Molar mass of barium is mathematically represented as

$Z = N_A V_a * \rho$

Where $N_A$ is the Avogadro's number

$N_A = \frac{ Z}{ V_a * \rho}$

substituting value

$N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}$

$N_A = 6.02289 *10^{23}$

4 0

3 years ago

Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the

hodyreva [135]

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

$amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}$

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

$amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}$