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Keith_Richards [23]
2 years ago
8

If you have 2.0 moles of sodium chloride (NaCl), what is its mass in grams?

Chemistry
1 answer:
lana [24]2 years ago
6 0

Answer:

117g

Explanation:

Given parameters:

Number of moles = 2moles

Unknown:

Mass of NaCl  = ?

Solution:

To solve the problem, we need to use the expression below;

    Mass of NaCl  = number of moles x molar mass

Molar mass of NaCl  = 23 + 35.5  = 58.5g/mol

 

So;

Insert the parameters and solve;

     Mass of NaCl  = 2 x 58.5  = 117g

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How many structures are possible for a square planar molecule with a formula of ax3y?
Galina-37 [17]

Answer:

Explanation:

Depending upon the relative arrangements of XandY X a n d Y , the square planar molecule AX3Y A X 3 Y shows only the following structure: Hence, only one structure is possible for a square planar molecule with a formula of AX3Y A X 3 Y .

5 0
2 years ago
Please I need help on these 3 questions. Thank You.​
Liula [17]

1.

V = 200 mL (volume)

c = 3 M = 3 mol/L (concentration)

First we convert mL to L:

200 mL = 0.2 L

Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol

Finally, we just use the molar mass of CaF2 to calculate the actual mass:

molar mass = 78 g/mol

The formula is: m = n × mm (mass = moles × molar mass)

m = 0.6 mol × 78 g/mol = 46.8 g

2.

For this question the steps are exactly like the first question.

V = 50mL = 0.05 L

c = 12 M = 12 mol/L

n = V × c = 0.05 L × 12 mol/L = 0.6 mol

molar mass (HCl) = 36.5 g/mol

m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.

3.

The steps for this question are the opposite way.

m(K2CO3) = 250 g

molar mass = 138 g/mol

n = m ÷ mm = 1.81 mol

c = 2 mol/L

V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL

6 0
3 years ago
If the ligand has a negative charge at a particular location, what would happen if you tried to put electrons from the metal nea
slega [8]

Answer:

The two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place

Explanation:

A complex is made up a central metal atom or ion and ligands. Ligands are lewis bases and they possess lone pairs of electrons. A complex is formed when electrons are donated from ligand species to metals.

However, if the ligand has a negative charge at a particular location and we try to put electrons from the metal near the electrons from the ligand, the two would end up repelling each other very strongly and more energy would ultimately be required to keep the metal-ligand system in place.

8 0
3 years ago
A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K
Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b)  Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

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