Answer:
because its not international language
Explanation:
because its not international language like English
Answer:
A) 0.20 cm³
B) 49.7 m²
C) 99.99%
D) 17.7 mg
Explanation:
A) The density of a material represents the mass that it occupies in a "piece" of volume. Thus, the density (d) is the mass (m) divided by the volume (v):
d =m/v
If the mass is 40.0 mg = 0.04 g, and the density is 0.20 g/cm³, the volume is:
0.20 = 0.04/v
v = 0.04/0.20
v = 0.20 cm³
B) The surface area (S) is the are that is presented in each gram of the material, so, it's the area (a) divided by the mass (m):
S = a/m
If the mass is 40.0 mg = 0.04 g, and the surface area is 1242 m²/g, so:
1242 = a/0.04
a = 49.7 m²
C) The percent of mercury removed is the mass removed divided by the initial mass, this multiplied by 100%. The mass removed is the initial mass (m0) less the final mass (m), so:
%removed = [(7.748 - 0.001)/7.748] *00%
%removed = 99.99%
D) The final mass of the spongy material is it mass (10 mg) plus the mass removed of the mercury (7.748 - 0.001 = 7.747 mg), so:
m = 10 + 7.747
m = 17.747 mg
m = 17.7 mg
The standard state of the elements Nitrogen and Oxygen are N2 and O2, knowing that they are diatomic elements. With that piece of information, the unbalanced equation for the formulation of NO2(g) should be as follows -
N2 + O2 ---> NO2
And if you include their states -
N2 ( g ) + O2 ( g ) ---> NO2 ( g )
To balance this chemical equation consider the number of reactants and products on other side of the equation. If you were to include a coefficient of one - half with respect to N2 on the reactant side, it would balance the reactants and products -
Answer: C. growth
Explanation: because why not
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