How many liters are in 2.94 mol HCl?

What is the volume of 45.6g of silver if the density of silver is 10.5g/mL? A. 4.34mL B. 479mL C. 0.23mL

nirvana33 [79]

Explanation:

first you get moles of silver

n=m/M

hence you add no of moles to this equation

c=nv

v=n/c

8 0

4 years ago

Why is gunshot residue important for a forensic scientist? What are the potential complications with this type of evidence?

erik [133]

A residue from a gunshot is most likely gun powder, which tells you what kind of bullet was shot and the type of gun that was used to shoot the target/victim/person. Some complications may be that there is more than one gun or weapon which uses that residue, so it may be hard to pinpoint it and the bullet can't really tell you who it is unless there's DNA on the bullet, and the chemicals of the bullet may even destroy evidence.

3 0

3 years ago

Read 2 more answers

Number of stable isotopes in Hydrogen?

kari74 [83]

3.
protium (A = 1), deuterium (A = 2), and tritium (A = 3).

5 0

3 years ago

Read 2 more answers

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$