Carbon Dioxide has two polar C=O. bonds, but the geometry of Carbon dioxide is linear so that the two bond dipole moments cancel and there is no net molecular dipole moment; the molecule is nonpolar.
I hope this helps :)
Answer:
0.9612 g
Explanation:
First we <u>calculate how many moles are there in 3.00 g of CCl₃F</u>, using its <em>molar mass</em>:
- 3.00 g CCl₃F ÷ 137.37 g/mol = 0.0218 mol CCl₃F
Now, we need to calculate how many grams of N₂O would have that same number of molecules, or in other words, <em>the same amount of moles</em>.
Thus we <u>calculate how many grams would 0.0218 moles of N₂O weigh</u>, using the <em>molar mass of N₂O</em> :
- 0.0218 mol N₂O * 44.013 g/mol = 0.9612 g N₂O
The time taken for the object to reach to top of pile is 0.012 year.
<h3>Time of motion </h3>
The time taken for the object to reach to top of pile is calculated as follows;
time of motion = distance traveled/speed
time of motion = (1.1 x 10¹⁴ x 10³ m)/(3 x 10⁸ m/s)
where;
- speed of light = 3 x 10⁸ m/s
time of motion = 3.67 x 10⁵ sec = 0.012 year
Thus, the time taken for the object to reach to top of pile is 0.012 year.
Learn more about time of motion here: brainly.com/question/2364404
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We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
Dry air is a mixture of nitrogen, oxygen, carbon dioxide etc.
air is a mixture of gases 78% nitrogen an 21% oxygen and other components.
