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EastWind [94]
3 years ago
10

Which mountain range was once the tallest in the world?

Chemistry
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

A

Explanation:

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A student combines 364.6 g of HCl with 80 g of NaOH in 5 L of water. What additional volume of H2O must be added to this mixture
Jobisdone [24]

Answer:

75L of additional water to have a pH 1 solution

Explanation:

The reaction of HCl With NaOH is:

HCl + NaOH → H₂O + NaCl

By using molar mass of each reactant you can know how many moles will react, thus:

HCl: 364.6g HCl ₓ (1mol / 36.46g) = 10 moles HCl

NaOH: 80g NaOH ₓ (1mol / 40g) = 2 moles NaOH

That means after the reaction will remain in solution, 10-2 = 8 moles of HCl = 8 moles of H⁺ (In water, HCl dissociates as H⁺ and Cl⁻ ions).

A solution with pH = 1 contains:

pH = -log [H⁺]

1 = -log [H⁺]

0.1M = [H⁺]

As molarity, M is the ratio between moles and liters and you want a solution 0.1M having 8 moles of H⁺ you require:

0.1M = 8 moles H⁺ / 80L

As the student combines the solution with 5L of water, you require

<h3>75L of additional water to have a pH 1 solution</h3>
6 0
3 years ago
How to do Lewis structure for Al?
Gnoma [55]

Notes:-

  • Al has Z=15

electron configuration:-

  • [Ne]3s²3p³

Valency is 3

So 3 dots are included

4 0
2 years ago
HS- is amphoteric; it can behave as either an acid or a base.
Tamiku [17]

Answer:

HS+Na=>NaS+1/2H2(here HS- acts as an acid)

HS-. + HCl=> H2S(g)+ Cl-(here HS- acts as a base)

7 0
3 years ago
analytical chemists can detect very small amounts of amoin acids , down to 3x10^-21 mol. how many molecules of an amino acid (mr
sesenic [268]
The number of particles in one mole is given be Avagadro's number <span>6.022×10^23

Multiply by number of moles.

3 ×10^-21 mol * 6.022 ×10^23 molecules/mol = </span><span>1,807 molecules
(rounded to nearest whole number)
 </span>
6 0
3 years ago
Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo
hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

q=mc\Delta T         .........(1)

where,

q = heat absorbed

m = mass of substance

c = specific heat capacity of substance

\Delta T = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

q=m\times \Delta H      ........(2)

where,

q = heat absorbed

m = mass of substance

\Delta H = enthalpy of the reaction

The processes involved in the given problem are:

1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K

Putting values in equation 1, we get:

q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J

  • <u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J

Total heat required = [q_1+q_2]

Total heat required = [4153.8J+32467.7J]=36621.5J

Hence, the total heat required is 36621.5 J

  • <u>For b:</u>

The processes involved in the given problem are:  

1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K

Putting values in equation 1, we get:

q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J

  • <u>For process 2:</u>

We are given:

m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g

Putting values in equation 2, we get:

q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J

  • <u>For process 3:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K

Putting values in equation 1, we get:

q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J

  • <u>For process 4:</u>

We are given:

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J

Total heat required = [q_1+q_2+q_3+q_4]

Total heat required = [448.14+4583.5+21445.2+32467.7]J=58944.5J

Hence, the total heat required is 58944.5 J

8 0
3 years ago
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