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Atomic mass of F: 19.0 g/mol
Atomic mass of S: 32.1 g/mol
1.18 g F = [1.18 g / 19.0 g / mol] = 0.062 mol F
1 g S = 1 g/ 32.1 g/mol = 0.031 mol S
Divide by 0.031
0.062 mol F / 0.031 = 2 mol F
0.031 mol S / 0.031 = 1 mol S
SF2 Then X = 2
Verification:
F2 = 2*19.0 g = 38 g F
S = 32.1 g
36 gF / 32.1 g S = 1.18 g F / g S
Answer:
32.711
Explanation:
1 yard = 0.0009144 km
1 week = 168 hrs
6.01 x 10^6 / 168 = 35,773.81 yds/hr
= 32.711 km/hr
Answer:
1.09 M
Explanation:
Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

Now, the total volume of this solution can be found by adding the volume values of each component:

Finally, dividing the moles found by the total volume will yield the final molarity:

Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.