Amount of CO₂ emission per day is 11,356.23 g.
<u>Explanation:</u>
Joe travelling distance per day = 60 miles
Carbon dioxide emission per day = 20 mpg
Now we have to find the amount of carbon dioxide emitted per day by dividing the distance by the emission per day given in gallons.
Amount of Carbon dioxide emission = 
Amount of CO₂ emission in gallons =
= 3 gallons
Now we have to convert the gallons to grams as,
1 gallon = 3,785.41 g
3 gallons = 3 × 3785.41 g = 11,356.23 g
So the emission of CO₂ per day is 11,356.23 g.
Answer:
8
Explanation:
From the question given above, the following data were obtained:
t–butyl ion = (CH₃)₃C⁺
Number of valence electron =?
The valence electron(s) talks about the combining power of an element or compound as the case may be.
Considering the t–butyl ion, (CH₃)₃C⁺ we can see that it has a charge of +1 indicating that it has given out 1 electron to attain the stable octet configuration which has a valence electrons of 8. Thus, the valence electron of t–butyl ion, (CH₃)₃C⁺ is 8
Atoms can be divided more so thats not true. atoms from the same element arent all nesacarly identical because of isotopes
Answer:
27.60 g urea
Explanation:
The <em>freezing-point depression</em> is expressed by the formula:
In this case,
- ΔT = 5.6 - (-0.9) = 6.5 °C
m is the molality of the urea solution in X (mol urea/kg of X)
First we<u> calculate the molality</u>:
- 6.5 °C = 7.78 °C kg·mol⁻¹ * m
Now we<u> calculate the moles of ure</u>a that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
- 0.84 m = mol Urea / 0.550 kg X
Finally we <u>calculate the mass of urea</u>, using its molecular weight:
- 0.46 mol * 60.06 g/mol = 27.60 g urea