<span>The water cycle has no starting point. But, we'll begin in the oceans, since that is where most of Earth's water exists</span>
Answer:
n = Initial volume/22.4L
Explanation:
The molar concept is simply one that is used to find the Number of moles and explain the relationship it has with avogadro's number, molecular mass, molar mass e.t.c.
Now, in terms of molar mass, number of moles is given by the formula;
n = mass of the sample/molar mass
In terms of avogadro's number, number of moles is;
1 mole = avogadro's number = 6.02 × 10^(23)
Now, when dealing with ideal gases, the molar volume of an ideal gas is 22.4 L.
Now the relationship between this volume and the mole concept is that the number of moles is gotten by dividing the initial volume by this molar volume.
Thus;
n = Initial volume/22.4L
The average speed is calculated using the following formula
Where
T = 298 K [room temperature]
m = mass of each molecule of oxygen
mass of one mole of oxygen molecules = 32 g / mol
mass of one molecule of oxygen will be = 
mass of one molecule of oxygen will be = 
putting values
average speed = 
Average speed = 482.19 m / s
average speed = 1735.8 km / h
so approx = 1700 km/h
Answer:
3 will be the correct coefficient of CaBr2
Explanation:
In balancing a chemical equation, numbers should be assigned to both reactants and products as a numerical coefficients until all atoms of elements in both sides of the equation count equal.
The balanced equation of the reaction will be:
3CaSO4 + 2AlBr3 ==> 3CaBr2 + Al2(SO4)3
Looking at the unbalanced equation in the question, in the product Al2(SO4)3 there are 3 SO4 group. This will warrant putting 3 behind CaSO4 in order to balance the atoms of SO4 group. That operation will automatically put the number of Ca atoms in CaSO4 to be 3 therefore making CaBr2 to have 3 coefficient as in the balanced equation. This is to balance the number of Ca atoms in both sides to be 3.
Answer:
There will be 143,67g CO2 produced
Explanation:
2 C6H6 + 15 O2 → 12 CO2 + 6 H2O
(42,5 g C6H6) / (78.1124 g C6H6/mol) = 0.54408775 mole C6H6
(113.1 g O2) / (31.9989 g O2/mol) = 3.534496 moles O2
0.54408775 mole of C6H6 would react completely with 0.54408775 x (15/2) = 4.080658 mole O2, but there is more O2 present than that, so O2 is in excess and C6H6 is the limiting reactant.
(0.54408775 mol C6H6) x (12/2) x (44.0096 g/mol) = 143.67 g CO2
