Answer:
6M
Explanation:
(Molarity x Volume)concentrated soln = (Molarity x Volume)diluted doln
Molarity dilute soln = [(M x V)conc/V (dilute)] = 1.5L x 12M / 3.0L = 6M final dilute soln
Answer:
energy required=qnet=87.75kJ
Explanation:
we will do it in three seperate step and then add up those value.
first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.
q 1= m c (T2-T1)
q1 = 36.0 g (4.18 J/gC) (100 - 65 C)
q1 = 5267 J
=5.267kJ
next is to vaporize the sample at 100C
q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol
q2= 81.4 kJ
Finally, heat the steam upto 115C
q3 = m c (T2-T1)
q 3= 36.0 g (2.01 J/gC)(115-100C)
q3 = 1085 J
=1.085kJ
qnet=q1 +q2 +q3
energy required=qnet=87.75kJ
Answer:
I would say the answer is D
Explanation:
A. The patch's area in square kilometers (km²) is 1.61×10⁻⁹ km²
B. The cost of the patch to the nearest cent is 734 cents
<h3>A. How to convert 16.1 cm² to square kilometers (km²)</h3>
We can convert 16.1 cm² to km² as illustrated below:
Conversion scale
1 cm² = 1×10⁻¹⁰ km²
Therefore,
16.1 cm² = 16.1 × 1×10⁻¹⁰
16.1 cm² = 1.61×10⁻⁹ km²
Thus, 16.1 cm² is equivalent to 1.61×10⁻⁹ km²
<h3>B. How to determine the cost in cent</h3>
We'll begin by converting 16.1 cm² to in². This can be obtained as illustrated below:
1 cm² = 0.155 in²
Therefore,
16.1 cm² = 16.1 × 0.155
16.1 cm² = 2.4955 in²
Finally, we shall the determine the cost in centas fo r llow:
- Cost per in² = $2.94 = 294 cent
- Cost of 2.4955 in² =?
1 in² = 294 cent
Therefore,
2.4955 in² = 2.4955 × 294
2.4955 in² = 734 cents
Thus, the cost of the patch is 734 cents
Learn more about conversion:
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It changes precipitation patterns, causes strong storms, and transfers warm and cold air.