Here are the choices:
A. the output current will be much smaller than the input current.
<span>B. the output voltage will be much larger than the input voltage. </span>
<span>C. the output voltage will be much smaller than the input voltage. </span>
<span>D. the output current will be much larger than the input current
</span>
The correct answer is letter <span>B. the output voltage will be much larger than the input voltage.</span>
a) 56g
<h3>Calculation:</h3>
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = 
= 
mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
Learn more about calculation at STP here:
brainly.com/question/9509278
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It would be HO2 Around there
Answer:
This question is incomplete
Explanation:
This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.
The formula to be used here is CₐVₐ/CbVb = nₐ/nb
where Cₐ is the concentration of the acid = unknown
Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)
Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)
Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)
nₐ is the number of moles of acid = 1 (from the chemical equation)
nb is the number of moles of base = 2 (from the chemical equation)
Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student
Cₐ x 25/0.105 x 22.13 = 1/2
Cₐ x 25 x 2 = 0.105 x 22.13 x 1
Cₐ x 50 = 0.105 x 22.13
Cₐ = 0.105 x 22.13/50
Cₐ = 0.047 mol/dm³
The concentration of the sulfuric acid is 0.047 mol/dm³
