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3 years ago

How many liters would you need to make a 0.8 M solution with 20 grams of lithium chloride?

Chemistry

1 answer:

worty [1.4K]3 years ago

5 0

Answer:

0.5875L

Explanation:

concentration = mole/ volume

n(LiCl) = 20 / (7 + 35.5) = 0.47 mol

volume = mole / conc.

volume = 0.47 /0.8

= 0.5875 dm³ = 0.5875L

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Help I. have until 11:59 to hand it in

Sedbober [7]

This chemical reaction can also be written as

2C2H6+7O2-->4CO2+6H2O

This is a chemical reaction because reactants are converted into products. This is the combustion of ethane where ethane and oxygen have been converted into carbon dioxide and water

8 0

3 years ago

A strong acid can also be described as a very weak .

faltersainse [42]

I guess is weak alkaline. when the substance is more acidic, there will be less alkalinity

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3 years ago

What are the roles of producers, primary consumers, secondary consumers, tertiary consumers for the trophic levels in the energy

Nesterboy [21]

This question is more for Biology than Chemistry, but the role of producers is to make energy (food) to be consumed. In a pyramid diagram, the producers would be at the bottom. Now going up the pyramid, the primary conumers are the first to consume producers and obtain energy from them. As you go up the pyramid, the secondary consumers will consume the primary consumers as a way to obtain energy, and the same goes for tertiary consumers towards secondaries.

As you go up the energy pyramid, you will notice a trend that there is less energy being obtained from each consumer. In other words, the producers will ALWAYS have more energy than the tertiary consumers.

I hope this answers your question.

3 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample

sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

Step 2: Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid * 26.38 kJ/g = 6.4631 kJ

Step 3: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ / 1.643°C = 3.934 kJ/ °C

Step 4: Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine

5.65 kJ / 0.0013389 moles = 4219.9 kJ/mol ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0

4 years ago