160gof CO2×1molof CO2\44g of CO2×1mole of C\1mol of CO2×12g of C\1mole of C=
43.63g of C
We can determine the empirical formula by first converting each of the grams to moles. remember to do this, first, we need the molar mass of the molecules which can be calculated by adding the mass of the atoms from the periodic table.
molar mass of CO2= 44.0 g/mol
molar mass of H2O= 18.02 g/mol
now, lets determine the grams of each atom
Carbon: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C
Hydrogen: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H
Oxygen: 10.0 - (6.54 + 0.55) = 2.91 g O
Now let's convert each mass to moles.
C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol
now that we have the moles of each atom, we need to divide them by the smallest value to find the ration. If you do not get the whole number, you need to multiply until to get a whole number.
C: 0.54 mol / 0.18 mol = 3
H: 0.54 mol / 0.18 mol = 3
O: 0.18 mol / 0.18 mol = 1
empirical formula--> C₃H₃O
Answer:
Chlorination
Explanation:
"Exposure to high volumes of chlorine gas fumes can cause serious health problems, including death."
https://water.mecc.edu/courses/ENV211/lesson14_print.htm
:)
Answer:
The hot tea should transfer <em>25.63 kJ</em> the surroundings to cool the tea.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat has to be transferred from the tea to the surroundings to cool the tea (Q = ??? J).
m is the mass of the hot tea (m = dV = (1.0 g/mL)(250 mL) = 250 g), suppose the density of water is the density of tea.
c is the specific heat of the hot tea (c = 4.10 J/°C.g).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 350 K - 375 K = -25°C).
<em>∴ Q = m.c.ΔT</em> = (250 g)(4.10 J/°C.g)(-25°C)) = <em>- 25630 J = - 25.63 kJ.</em>
<em>So, the hot tea should transfer 25.63 kJ the surroundings to cool the tea.</em>
