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STatiana [176]
3 years ago
11

A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the

volume of the solvent does not change when the aniline dissolves in it.
Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.

How do I enter a number in scientific notation?

a. molarity =
b. molality =
Chemistry
1 answer:
Tresset [83]3 years ago
7 0

Answer:

a. Molarity= M =2.1x10^{-1}M

b. Molality= m=2.0x10^{-1}m

Explanation:

Hello,

In this case, given the information about the aniline, whose molar mass is 93g/mol, one could assume the volume of the solution is just 200 mL (0.200 L) as no volume change is observed when mixing, therefore, the molarity results:

M=\frac{n_{solute}}{V_{solution}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L} =2.1x10^{-1}M

Moreover, the molality:

m=\frac{n_{solute}}{m_{solvent}} =\frac{3.9g*\frac{1mol}{93g} }{0.2L*\frac{1.05kg}{1L} } =2.0x10^{-1}m

Best regards.

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How does the volume of water change the solubility of sodium chloride in simple words?
kupik [55]

Answer:

When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. ... The solubility of salt does not change much with temperature, so there is little profit in using hot water.

3 0
3 years ago
How many atoms of hydrogen are present in 7.63 g of ammonia? please help me ??
GarryVolchara [31]
The atoms  of  hydrogen  that are  present  in  7.63 g  of ammonia(NH3)

find  the  moles  of NH3 =mass/molar mass
 7.63 g/ 17 g/mol = 0.449  moles

since there  is  3 atoms of H  in NH3 the  moles of  H = 0.449 x 3 = 1.347 moles

by  use  of  1 mole = 6.02 x10^23  atoms
what  about  1.347  moles

= 1.347  moles/1   moles  x 6.02 x10^23 atoms = 8.11 x10^23  atoms of Hydrogen
3 0
3 years ago
Water flows over Niagara Falls at the average rate of 2,400,000 kg/s, and the average height of the falls is about 50 m. Knowing
Zinaida [17]

Answer:

1) The power of Niagara Falls is 1.176 × 10⁹ W

2) The number of 15 W LED light bulbs it could power is 78.4 × 10⁶ light bulbs

Explanation:

1) The Niagara falls water mass flow rate = 2,400,000 kg/s

The height of the fall = 50 meters

The gravitational potential energy = Mass (kg) × height (m) × gravity (9.8 m/s²)

The power = The energy converted per second = Mass flow rate (kg/s) × height (m) × gravity (9.8 m/s²)

Therefore;

The power of Niagara Falls= 2,400,000 kg/s × 50 m ×9.8 m/s²= 1.176 × 10⁹ W

The power of Niagara Falls = 1.176 × 10⁹ W

2) The number, n, of 15 W LED light bulbs it could power is given by the relation;

n × 15 W = 1.176 × 10⁹ W

∴ n = 1.176 × 10⁹ W/(15 W) = 78.4 × 10⁶ light bulbs

The number of 15 W LED light bulbs it could power = 78.4 × 10⁶ light bulbs.

6 0
3 years ago
Use the value of the Avogadro constant (6.02 × 1023 mol–1) to calculate the total number of atoms in 7.10 g of chlorine atoms. (
kolbaska11 [484]

The total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.

HOW TO CALCULATE NUMBER OF ATOMS:

  • The number of atoms in a substance can be calculated by multiplying the number of moles in that substance by Avogadro's number as follows:

  • no. of atoms = no. of moles × 6.02 × 10²³ mol-¹

  • The number of moles in 7.10g of Cl is calculated as follows:

no. of moles = mass ÷ molar mass

no. of moles = 7.10g ÷ 35.5g/mol

no. of moles = 0.2mol

no of atoms = 0.2mol × 6.02 × 10²³

no. of atoms = 1.204 × 10²³atoms.

  • Therefore, the total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.

Learn more: brainly.com/question/15488332?referrer=searchResults

6 0
2 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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