Answer:
82 days would take a distance like that, if we drive at 120 mi/h
Explanation:
Distance from Earth to moon is 3.8×10⁵ km
You drive at a speed of 120 mi/h
Let's convert the 120 mi/h to km/h
1 mi = 1.61 km
Then 120 mi . 1.61 km / 1mi = 193.2 km
New velocity is 193.2 km. Therefore,
193.2 km are driven in 1 hour
3.8×10⁵ km woul be driven in (3.8×10⁵ . 1)/ 193.2 = 1966.8 hours
1967 hours . 1 day / 24 hours = 82 days
The equation for the reaction between NaOH and AlCl₃ is as follows;
3NaOH + AlCl₃ ---> 3NaCl + Al(OH)₃
the stoichiometry of NaOH : AlCl₃ is 3:1
3 moles of NaOH reacts with 1 mol of AlCl₃ to produce 1 mol of Al(OH)₃
the number of AlCl₃ moles reacted - 6.5 mol
molar mass of NaOH -(23 +16 +1) = 40 g/mol
the number of NaOH moles reacted = 57.0 g / 40 g/mol
NaOH moles = 1.425 mol
either NaOH or AlCl₃ is in excess and other is the limiting reactant.
limiting reactant is the reactant whose number of moles are fully consumed during the reaction. the reactant that is in excess will have leftover moles that are remaining after the reaction.
If AlCl₃ is the limiting reactant, number of NaOH moles would be thrice the amount of AlCl₃ present,
then number of NaOH moles that should be present - 6.5 * 3 = 19.5 mol
however there are only 1.425 mol of NaOH present, therefore AlCl₃ is in excess.
Then NaOH is the limiting reactant,
the amount of products formed depends on the amount of the limiting reactant present.
stoichiometry of NaOH : Al(OH)₃ is 3:1
the number of Al(OH)₃ moles produced = number of NaOH moles reacted / 3
number of Al(OH)₃ moles are - 1.425 mol /3 = 0.475 mol
molar mass of Al(OH)₃ = (27 +3*16 + 3*1) = 78 g/mol
mass of Al(OH)₃ produced = 78 g/mol * 0.475 mol = 37.05 g
Correct Question :
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80°C
In an experiment to determine the specific heat of a metal student transferred a sample of the metal that was heated in boiling water into room temperature water in an insulated cup. The student recorded the temperature of the water after thermal equilibrium was reached. The data we shown in the table above. Based on the data, what is the calculated heat absorbed by the water reported with the appropriate number of significant figures?
Answer:
1642 J
Explanation:
Given:
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80° C
To calculate the heat absorbed by water, Q, let's use the formula :
Q = ∆T * mass of water * specific heat
Where ∆T = 32.80°C - 24.95°C = 7.85°C
Therefore,
Q= 7.85 * 50.003 * 4.184
Q = 1642.32 J
≈ 1642 J
Answer:
if a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one-half
