Write the electron configuration for the element titanium, Ti.
1 answer:
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Answer:
3) About 0.35 grams of hydrogen gas.
4) About 65.2 grams of aluminum oxide.
Explanation:
Question 3)
We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.
Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:
To balance it, we can simply add another sodium atom on the left. Hence:
To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.
The molar mass of sodium is 22.990 g/mol. Hence:
From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:
And the molar mass of hydrogen gas is 2.016 g/mol. Hence:
Given the initial value and the above ratios, this yields:
Cancel like units:
Multiply. Hence:
Since we should have two significant values:
So, about 0.35 grams of hydrogen gas will be released.
Question 4)
Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:
To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:
To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.
The molar mass of aluminum is 26.982 g/mol. Thus:
According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:
And the molar mass of aluminum oxide is 101.961 g/mol. Hence:
Using the given value and the above ratios, we acquire:
Cancel like units:
Multiply:
Since the resulting value should have three significant figures:
So, approximately 65.2 grams of aluminum oxide is produced.
i. The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.
ii. The structural formula of Y is and Z is
.
iii. Alkenes, upon catalytic hydrogenation, form alkanes. This will occur in the presence of Nickel as the catalyst.
iv. The process of ozonolysis is useful in the field of pharmaceutics.
v. The test of unsaturation can be performed by passing a compound through Bromine water.
<h3>What is ozonolysis?</h3>Ozonolysis is a reaction used in organic chemistry to determine the position of a carbon-carbon double bond in unsaturated compounds.
i. The given alkene X, that is subject to ozonolysis would be 2-methyl-2-pentene. Upon exposure to ozone, an ozonide is initially formed, after which it is broken down into 2 products - acetone and propanal, both with the molecular formula C₃H₆O.
The given compound X is 2-methyl pent-2-ene. When it is reacted with ozone it forms an ozonide in the first step. In the second step the reduce to forms acetone and propanal.
ii. The formulas of Y is and Z is
. They are functional isomers as they have the same molecular formula but different functional groups - ketone and aldehyde.
iii. When alkenes undergo catalytic hydrogenation, they form alkanes, X will form 2 methyl petane on reaction with hydrogen gas in presence of Ni.
iv. The ozonolysis is used for the industrial-scale synthesis of pharmaceuticals.
v. The unsaturation of compound X can be proved by the bromine water test. As on reaction with it, the brown colour of bromine water becomes colourless due to the formation of dibromo alkane.
Learn more about the ozonolysis here:
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