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3 years ago

Identify each process numbered 1-6

Chemistry

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

1 = Melting

2 = Freezing

3 = Sublimation

4 = Deposition

5 = Condensation

6 = Boiling

Explanation:

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. Calculate the molality of each of the following solutions: (a) 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg

Step2247 [10]

Answer:

a)0.67 molal

b) 5.67 molal

c) 2.82 molal

d) 0.0268 molal

Explanation:

Step 1: Data given

Molality = moles solute / mass solvent

(a) 0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C

Calculate moles of Na2CO3 = 710 grams / 105.99 g/mol = 6.70 moles

Molality = 6.70 moles / 10 kg = 0.67 molal

(b) 125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack

Calculate moles of NH4NO3 = 125 grams / 80.04 g/mol = 1.56 moles

molality = 1.56 moles / 0.275 kg = 5.67 molal

(c) 25 g of Cl2 in 125 g of dichloromethane, CH2Cl2

Calculate moles Cl2 = 25 grams / 70.9 g/mol= 0.353 moles

Molality = 0.353 moles / 0.125 kg = 2.82 molal

0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3

Calculate moles histamine = 0.372 grams / 111.15 g/moles = 0.00335 moles

Calculate molality = 0.00335 moles / 0.125 kg = 0.0268 molal

NOTE: histamine is C5H9N3 not C5H9N

4 0

3 years ago

A 6.0 Liter solution is made with 2 moles of solute. What is the molarity of the solution?

kodGreya [7K]

The formula for molarity is: mol/L. so it would be 2.0 mol/ 6.0 L.

the answer is: .33 M

hope this helps!

3 0

3 years ago

A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0

3 years ago

While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°

Lana71 [14]

Answer:

Final temperature = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C

305 cal = 5.25cal/°C × T2 - 25°C

305 cal / 5.25cal/°C = T2 - 25°C

58.1 °C = T2 - 25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

7 0

2 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago