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2 years ago

In general, what makes one kind of amino acid different from other amino acids?

1 answer:

5 0

There is part of an amino acid molecule that is called the R group or side chain. The side chain of the amino acid called glycine is a single hydrogen atom. The side chain is what differs from amino acid to amino acid.

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Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th

Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ = C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

6 0

2 years ago

What characteristics does the serval have that make it well-suited to life on the savanna ?

nika2105 [10]

It is very camouflaged this way it will survive longer then most other animals

5 0

3 years ago

Read 2 more answers

Antibiotics do not reduce the number of helpful and harmful bacteria in the microbiome.

stira [4]

Answer:

false; the purpose of antibiotics is to kill bacteria

Explanation:

7 0

3 years ago

Read 2 more answers

A 13.00 g sample of citric acid reacts with an excess of baking soda What is the theoretical yield of carbon dioxide

viva [34]

Answer:

8.934 g

Step-by-step explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 192.12 44.01

H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂

m/g: 13.00

For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.

(a) Calculate the moles of H₃Cit

n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)

n = 0.067 67 mol H₃Cit

(b) Calculate the moles of CO₂

The molar ratio is (3 mol CO₂/1 mol H₃Cit)

n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)

n = 0.2030 mol CO₂

(c) Calculate the mass of CO₂

m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)

m = 8.934 g CO₂

4 0

3 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

For $N_2$ :

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

For $H_2$ :

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

2 years ago