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3 years ago

The conversion of large hydrocarbon molecules into smaller molecules is known as:

Chemistry

1 answer:

Lerok [7]3 years ago

6 0

Answer:

Thermal cracking, also known as visbreaking, is an older process that capitalizes on heat and pressure to break large hydrocarbon molecules into smaller, light molecules. The more-modern and more-efficient technology is catalytic cracking.

Explanation:

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Select the best definition of salt water:

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Water that contains large amounts of disolved salts

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2 years ago

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A variety of species of Galapagos finches evolved from one original species long ago through the process of

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Answer:

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Explanation:

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3 years ago

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Ca(NO3)2 + HCl --> HNO3 + CaCl2

aliina [53]

Calcium nitrate is limiting reactant, since it says there’s excess HCl. So you must use the mass of calcium nitrate in your calculation (see picture below):

5 0

3 years ago

What is the final [Na+] in a solution prepared by mixing 70.0 mL of 3.00 M Na2SO4 with 30.0 mL of 1.00 M NaCl?

Lynna [10]

Answer:

4.5 M

Explanation:

70.0 ml was mixed in 3.00 M of Na2SO4

30.0 ml was mixed in 1.00 M of NACL

The first step is to convert 70 ml to liters

= 70/1000

= 0.07 liters

The formular for molarity is

moles/liters

The number of moles in Na2S04 can be calculated as follows

Let y represent the number of moles

3M= y moles/0.07

= 3×0.07

= 0.21 moles

Since Na2So4 has 2 moles of Na then the number of moles is

= 2×0.21

= 0.42 moles

Convert 30ml to liters

= 30/1000

= 0.03 liters

The number of moles in Nacl can be calculated as follows

Let y represent the number of moles

1M= y moles/0.03

= 1×0.03

= 0.03 moles

Since Nacl has 1 mole of Na then the number of moles is

= 1 × 0.03

= 0.03 moles

Therefore the final Na+ can be calculated as follows

Total moles = 0.03 moles + 0.42 moles

= 0.45 moles

Total liters= 0.07 liters + 0.03 liters

= 0.1 liters

Na+ = 0.45/0.1

= 4.5 M

Hence the final Na+ in the solution is 4.5 M

7 0

4 years ago

A 0.2500 g sample of an alloy reacts with to form hydrogen gas: 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g) Zn(s) + 2H+(aq) Zn2+(aq) + H

fenix001 [56]

Answer: The mass percent of zinc in the alloy is 78.68 %

Explanation:

We are given:

Mass of sample of alloy = 0.2500 g

Let the mass of aluminium be 'x' grams and mass of zinc will be (0.2500 - x) g

To calculate the amount of hydrogen gas produced, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 0.147 L

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 0.147L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{755\times 0.147}{62.3637\times 298}=0.00597mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For Aluminium:

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{x}{27}mol$

The chemical equation follows:

$2Al(s)+6H^+(aq.)\rightarrow 2Al^{3+}(aq.)+3H_2(g)$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, $\frac{x}{27}$ moles of aluminium will produce = $\frac{3}{2}\times \frac{x}{27}=\frac{3x}{54}mol$ of hydrogen gas

For Zinc:

Molar mass of zinc = 65.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of zinc}=\frac{(0.25-x)}{65.4}mol$

The chemical equation follows:

$Zn(s)+2H^+(aq.)\rightarrow Zn^{2+}(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of zinc produces 1 moles of hydrogen gas

So, $\frac{(0.25-x)}{65.4}$ moles of zinc will produce = $\frac{1}{1}\times \frac{(0.25-x)}{65.4}=\frac{(0.25-x)}{65.4}mol$ of hydrogen gas

Equating the moles of hydrogen gas:

$\Rightarrow 0.00597=\frac{3x}{54}+\frac{(0.25-x)}{65.4}\\\\x=0.0533g$

To calculate the mass percentage of zinc in alloy, we use the equation:

$\text{Mass percent of zinc}=\frac{\text{Mass of zinc}}{\text{Mass of alloy}}\times 100$

Mass of zinc = (0.2500 - x) = [0.2500 - 0.0533] = 0.1967 g

Mass of alloy = 0.2500 g

Putting values in above equation, we get:

$\text{Mass percent of zinc in alloy}=\frac{0.1967g}{0.2500g}\times 100=78.68\%$

Hence, the mass percent of zinc in the alloy is 78.68 %

8 0

4 years ago