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.
Answer:
Investigating a data breach
Explanation:
A data breach usually involves data exfiltration over a computer network. the other options involve data being stored on a device locally which isn't volatile data like text messages, photos or rearranging data in defragmentation all of which does not require a network.
Answer:
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Explanation:
Mass of the solution = m
Mass of the KCN in solution = 779 mg
Mass by mass percentage of KCN solution = 0.173%
1 mg = 0.001 g
m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g
450. g of 0.173 % KCN solution contains 779 mg of KCN.
A hydrogen bond<span> is the electrostatic attraction between two polar groups that occurs when a </span>hydrogen<span> (H) atom covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative atom nearby. examples h20</span>
Answer:
Second element(Titanium); [Ar] 3d2 4s2
Third element(Vanadium):Ar 3d3 4s2
Explanation:
Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;
Second element(Titanium); [Ar] 3d2 4s2
Third transition element(Vanadium):Ar 3d3 4s2
Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.
