Mass of solute ( m1 ) = 50.0 g
mass of solvent ( m2 ) = 150.0 g
Therefore:
m/m = ( m1 / m1 + m2 )
m/m = ( 50.0 / 50.0 + 150.0 )
m/m = ( 50.0 / 200 )
m/m = 0.25
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Ocean currents determine the various directions of wind movement.
<h3>What is ocean current?</h3>
Ocean currents are caused by differences in density as well as the temperature of the moving winds across the ocean. In the ocean, warm water is found at the top while cooler water occurs far below.
Warm ocean currents originate near the equator and move towards the poles. The ocean currents control the direction of winds in an area.
Learn more about ocean currents: brainly.com/question/20823678
(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???
when the reaction equation is:
C(s) + H2O(g) ↔ H2(g) + CO(g)
∴ Kc = [H2] [CO] / [H2O]
and we have Kc = 0.0393 (given missing in the question)
when the O2 is added so, the reaction will be:
2H2(g) + O2(g) → 2H2O(g)
that means that 0.15 mol H2 gives 0.15 mol of H2O
∴ by using ICE table:
[H2O] [H2] [CO]
initial 0.57 + 0.15 0 0.15
change -X +X +X
Equ (0.72-X) X (0.15+X)
by substitution:
0.0393 = X (0.15+X) / (0.72-X) by solving for X
∴ X = 0.098
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
= 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
= 0.72 - 0.098
= 0.622 M
