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3 years ago

Determine how many mm of Hg are equal to 5.3 atm of pressure

1 answer:

5 0

The answer is 4,028.0 mm of Hg

It is known that:
1 atm of pressure = 760.0 mm of Hg

Hence:
5.3 atm of pressure = x mm of Hg

Let's use the proportion:
1 atm : 760.0 mm Hg = 5.3 atm : x

After crossing the products:
x = 760.0 mm Hg * 5.3 atm : 1 atm
x = 4,028.0 mm of Hg

Therefore, 4,028.0 mm of Hg are equal to 5.3 atm of pressure

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The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

Alenkinab [10]

Answer:

2.943 °C temperature change from the combustion of the glucose has been taken place.

Explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.590 g

Moles of benzoic acid = $\frac{0.590 g}{122.12 g/mol}=0.004831 mol$

Energy released by 0.004831 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004831 mol=15.5955 kJ=15,595.5 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.125°C

$Q=C\times \Delta T$

$15,595.5 J=C\times 2.125^oC$

$C=7,339.05 J/^oC$

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose = $\frac{1.400 g}{180.16 g/mol}=0.007771 mol$

Energy released by the 0.007771 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.007771 mol=21.6030 kJ=21,603.01 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$21,603.01 J=7,339.05 J/^oC\times \Delta T'$

$\Delta T'=2.943^oC$

2.943 °C temperature change from the combustion of the glucose has been taken place.

8 0

3 years ago

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

A the absorbance,
$l$ the path length,
ε the molar absorptivity of the solute, and
$c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

What is creation in terms of chemical reactions?

shtirl [24]

Answer:

Products

Explanation:

In a chemical reaction, the atoms and molecules produced by the reaction are called products

7 0

3 years ago

As the Earth formed, naturally occurring radioactive material _____________.

lesya692 [45]

Naturally Occurring Radioactive Materials (also known as NORM) are a wide range of radioactive isotopes that include elements such as carbon 14 and potassium 40, both of which are in the human body. But the main radioactive elements involved in oil and gas production are those found throughout Earth's crust. These elements include uranium and thorium and their respective byproducts, including radon gas.

6 0

3 years ago

Read 2 more answers

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

aalyn [17]

Answer:

2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, $Me_{2}S$ etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Reaction steps are shown below.

8 0

3 years ago