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3 years ago

How many moles are in 19.6 g of Sodium (Na)? And the conversion factor?

Chemistry

1 answer:

Snezhnost [94]3 years ago

6 0

Answer:

n=0.852 moles

Explanation:

Given mass is, m = 19.6 g

The molar mass of sodium is, M = 22.99 u

We need to find the no of moles in 19.6 of Sodium. We know that, no of moles is equal to given mass divided by molar mass.

$n=\dfrac{m}{M}\\\\n=\dfrac{19.6}{22.99}\\\\n=0.852$

So, there are 0.852 moles in 19.6 g of Sodium.

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2 years ago

Can sedimentation and decantation be used for all types of mixtures? Explain

Bumek [7]

Explanation:

1. Sedimentation and decantation cannot be used for all types of mixtures.

Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.

In decantation, gravity is used to bring the denser materials to settle at the bottom.

For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.

2. Yes, mass of an object reduces the settling time of such object in a mixture.

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2 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

5 0

2 years ago

Enter your answer in the provided box. From the data below, calculate the total heat (in J) needed to convert 0.304 mol of gaseo

diamong [38]

Answer:

-35,281.5 J

Explanation:

To convert the gaseous ethanol to liquid ethanol, three steps will occur. First, it will lose heat and the temperature will decrease until its boiling point, so from 300.0°C to 78.5°C. Thus, more heat will be lost, but now, with the temperature constant, so the gas will be converted to liquid. And then, the liquid will lose heat to decrease the temperature from 78.5°C to 25.0°C.

The total heat loss is the sum of the heats of each step. Because the heat is being removed from the system, it's negative. The first and last step occurs with a change in temperature, and so the heat is calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat of the gas (first step) or liquid (last step), and ΔT the temperature variation (final - initial). The mass of ethanol is the molar mass 46.07 g/mol multiplied by the number of moles, so:

m = 46.07 * 0.304 = 14.00 g

The second step occurs without a change in temperature, and the heat is then:

Q = -n*ΔH°vap

Where n is the number of moles, ΔH°vap is the heat of vaporization, and the minus signal indicates that the heat is being lost. Then, the heat of each step is:

Q1 = 14.00*1.43*(78.5 - 300,0) = -4434.43 J

Q2 = -0.304*40.5 = -12.312 kJ = -12312 J

Q3 = 14.00*2.45*(25.0 - 78.5) = -1835.05 J

Q = Q1 + Q2 + Q3

Q = -35,281.5 J

5 0

3 years ago