The correct answer is A. C3H4 is the three-carbon alkene contains two double bonds. Alkenes are a class of organic compounds with carbon-carbon double bond. The structure for C3H4 is:
<span>
H2C=C=CH2</span>
Percentage
by mass is the amount in mass of a component in a mixture per 100 unit of mass of the total
mixture. Percentage by mass is the same as %w/w. In order to determine this value, we need to know the mass of the substance in question and the total mass of the sample. In this case, we have to know the mass of Fe by doing certain calculations. Then, we divide this value by the total mass given of the ore, 4.05 g and multiply by 100%. It should be expressed as follows:
Percent by mass = mass of Fe in ore / 4.05 g ore sample x 100
However, the question is lacking some information so we cannot really specify the answer.
<em><u>Answer </u></em><em><u>:</u></em><em><u> </u></em><em><u>M</u></em><em><u>etal containers</u></em><em><u> </u></em><em><u>are</u></em><em><u> not</u></em><em><u> </u></em><em><u> </u></em><em><u>used for storing acid</u></em><em><u> because most of the time acid reacts with almost every metal and produces </u></em><em><u>salts</u></em><em><u> or oxid</u></em><em><u>e</u></em><em><u>s</u></em><em><u> </u></em><em><u>which alters the acid characteristics making it useless</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
i dont understand anything you typed sorry
Explanation:
Answer:
99 L
Explanation:
Step 1: Given data
Moles of nitrogen: 4.4 moles
Step 2: Calculate the volume occupied by 4.4 moles of nitrogen
The volume occupied by a gas depends on other conditions such as pressure (P) and temperature (T). If we have this information, we can calculate the volume of the gas using the ideal gas condition.
P × V = n × R × T
Since the task doesn't inform the conditions, we can assume it is under standard pressure (1 atm) and temperature (273.15 K). At STP, 1 mole of any gas occupies 22.4 L.
4.4 mol × 22.4 L/1 mol = 99 L
