Answer:
Explanation:
Given parameters:
Mass of CuCl₂ = 2.50g
Mass of Al = 0.50g
Unknown:
Number of moles of CuCl₂ and Al = ?
Solution:
To solve this problem, we must understand that the number of moles is a fundamental property used in stoichiometry calculations.
Number of moles = 
Molar mass of CuCl₂ = 63.6 + 2(35.5) = 134.5g/mole
Molar mass of Al = 26.98g/mole
Number of moles of CuCl₂ = 
= 0.019moles
Number of moles of Al = 
= 0.019moles
Answer:
- <u>1. Equation: 2x + 3 = 9x - 11</u>
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- <u>2. Each row has 2 chairs</u>
Explanation:
The variable x represents the number of chairs in each row.
<u />
<u>1. She can form 2 rows of a given length with 3 chairs left over.</u>
Thus, she has:
number of rows number of chairs in chairs number of chairs
each row left over she has
2 x 3 2x + 3
<u>2. She can form 9 rows of the same length if she gets 11 more chairs.</u>
That means that she is short in 11 chairs to have 9x chairs, or that she has 11 less chairs than 9x chairs. Then she has:
<u>3. Equation:</u>
Then, number of chairs she has is 2x + 3 and, also, 9x - 11, which allows to set the equation:
<u>4. Solve the equation:</u>
Therefore, each row has 2 chairs, and she has 2x + 3 = 4 + 3 = 7 chairs.
Answer: The beaker will not tip over when placed on the hot plate
Justification:
Since beakers have flat surface bottoms (usually and this is the condition to use them for this particular application) they can be placed safely on the hot plate without the risk that the they tip over.
Beakers are wide mouth cylindrical vessels used in laboratories to store, mix and heat liquids. Most are made of glass, in which case the glass is resistant to the flame and does not break when exposed to high temperatures or when is heated by direct contact on a hot plate.
So, their safe shape (flat bottom) that makes them stable, along with their ability to withstand high temperatures, make them suitable to heat solutions in laboratories.
We use Charles's Law: V1/T1=V2/T2
Standard Temperature: 0 degree Celsius= 273K
333.0 degrees Celsius= 606K
Set up: (1.00L)/ (273K)= V2/ (606.0K)
⇒ V2= (1.00L)/ (273K)* (606.0K)= 2.22L
Hope this would help :))
<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g
