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1 year ago

2. A chemical analysis of a sample provides the following elemental data:

Chemistry

1 answer:

Vadim26 [7]1 year ago

3 0

Answer:

C3 H6 O2

Explanation:

first divide their mass by their respective molar mass, we get:

30.4 moles of C

61.2 moles of H

20.25 moles of O

now divide everyone by the smallest one of them then we get

C= 1.5

H= 3

O= 1

since our answer of C is not near to any whole number so we will multiply all of them by 2

so,

C3 H6 O2 is our answer

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Identify the reactants and the products of the preparation step to the urea cycle.

Salsk061 [2.6K]

According to sources, the most probable answer to this query is the enzymes and waste products that are collected by the nephron from the blood. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

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2 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

What is the new concentration of 25.0mL added to 125.0mL of 0.150M

____ [38]

Answer:

The new concentration is 0.125 M.

Explanation:

Given data:

Initial volume V₁ = 125.0 mL

Initial molarity M₁ = 0.150 M

New volume V₂ = 25 mL +125 mL = 150 mL

New concentration M₂ = ?

Solution:

M₁V₁ = M₂V₂

0.150 M × 125 mL = M₂ × 150 mL

M₂ = 0.150 M × 125 mL / 150mL

M₂ = 18.75 M.mL/150 mL

M₂ = 0.125 M

The new concentration is 0.125 M.

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3 years ago

What is the answer to this

scoray [572]

The answer is 128.19

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3 years ago

Phase changes!a. Write out the Clapeyron equation and explain what it describes.b. Write out the Clausius-Clapeyron equation and

djverab [1.8K]

Answer:

Solutions in the attachment.

Explanation:

4 0

2 years ago