I just took a test with this question and got the answer wrong for saying ethane. The correct answer is propane.
Velocity = 3.61 m/s
<h3>Further explanation</h3>
Given
mass of child = 35 kg
mass of sled = 5 kg
Kinetic energy = 260 J
Required
velocity
Solution
Energy because its motion is expressed as Kinetic energy (KE) which can be formulated as:
KE = 1/2.mv²
mass of object :
= mass child + mass sled
= 35 kg + 5 kg
= 40 kg
Input the value :
v²=KE : 1/2.m
v²= 260 : 1/2.40 kg
v²=13
v=3.61 m/s
<span>the balanced equation for the reaction is as follows
Na</span>₂<span>SO</span>₄<span> + BaCl</span>₂<span> ----> 2NaCl + BaSO</span>₄
<span>stoichiometry of Na</span>₂<span>SO</span>₄<span> to BaCl</span>₂<span> is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl</span>₂<span> to BaSO</span>₄<span> is 1:1.
therefore number of BaSO4 moles formed - 0.288 mol</span>
Now ,
C + O2 → CO2
According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.
No of moles = mass of the substance/molecular mass of the substance.
In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.
No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.
No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles
Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.
Also number of moles of O2 = mass of O2÷ molar mass of O2
Substituting number of moles of O2 as 0.1 we get
mass of O2(x) = Number of moles of O2 × Molar mass of O2
Mass of O2 (x) = 0.1 × 32= 3.2 g
Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.
