Answer:
It would take 64 1-cm cubes to construct a 4-cm cube.
Explanation:
The volume of a cube is equal to the edge raised to the power of 3.
A 1-cm edge cube has a volume of V₁ = (1 cm)³= 1 cm³
A 4-cm edge cube has a volume of V₄ = (4 cm)³= 64 cm³
The number of small cubes required to fill the same volume than the big cube is:
Answer:
Explanation:
Given parameters:
Mass of MgCl₂ = 2.52 x 10⁻⁴g
Volume of water = 2L
Unkown:
Concentration of MgCl₂ =?
Concentration of Mg²⁺ = ?
Concentration of Cl⁻ =?
Solution:
Concentration is defined as the number of moles of a solute contained in a solution.
Concentration =
To find the number of moles"
number of moles =
Molar mass of MgCl₂ = 24.3 + (2 x 35.5) = 95.3g/mol
number of moles = = 0.000252moles
Concentration of MgCl₂ = = 2.64 x 10⁻⁶moldm⁻³
from the formula of the compound;
1 mole of MgCl₂ contains 1 mole of Mg²⁺
Therefore, 2.64 x 10⁻⁶moldm⁻³ of MgCl₂ will contains 2.64 x 10⁻⁶moldm⁻³ of Mg
Also;
1 mole of MgCl₂ contains 2 mole of Cl⁻
2.64 x 10⁻⁶moldm⁻³ contains 2 x 2.64 x 10⁻⁶moldm⁻³; 5.29 x 10⁻⁶moldm⁻³
Expressing in ppm;
1ppm = 1mg/L
2.64 x 10⁻⁶moldm⁻³ to mg/L for Mg²⁺
2.64 x 10⁻⁶moldm⁻³ = 2.64 x 10⁻⁶moldm⁻³ x molar mass(g/mol)
= 2.64 x 10⁻⁶moldm⁻³ x 24.3 = 6.43 x 10⁻⁵g/L
g/L to mg/L; 6.43 x 10⁻⁵g/L x 1000 = 6.43 x 10⁻²mg/L = 6.43 x 10⁻²ppm
5.29 x 10⁻⁶moldm⁻³ to mg/L for Cl⁻;
5.29 x 10⁻⁶moldm⁻³ = 5.29 x 10⁻⁶moldm⁻³ x 35.5 = 1.88 x 10⁻⁴g/L
g/L to mg/L; 1.88 x 10⁻⁴g/L x 1000 = 1.88 x 10⁻¹mg/L = 1.88 x 10⁻¹ppm
Answer:
40.02 calories
Explanation:
V = 10 mL = 10g
we know t went <em>up</em> by 4°C, this is our ∆t as it is a change.
Formula that ties it together: Q = mc∆t
where,
Q = energy absorbed by water
m = mass of water
c = specific heat of water (constant)
∆t = temperature change
Q = (10 g) x (4.186 J/g•°C) x (4°C)
Q = 167.44 J
Joules to Calories:
167.44 J x 1 cal/4.184 J = 40.02 calories
(makes sense as in image it is close to the value).
