Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g

1 answer:

4 0

Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

$gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl$

Have a nice day!

You might be interested in

What is the chemical formula for a compound between Li and Br?

natima [27]

LiBr.

<h3>Explanation</h3>

Note that the group number in this answer refers to the new IUPAC group number, which ranges from 1 to 18. Counts from the left. Start with the first two column (group 1 and 2), go on to the transition elements (Sc, Ti, etc. in group 3 through 12), and continue with the nonmetals (group 13 through 18).

Li is a group 1 metal. As a metal, it tends to form positive ions ("cations"). Metals in group 1 and 2 are <em>main group</em> metals. The charge on main group metal ions tends to be the same as the group number of the metal. Li is in group 1. The charge on an Li ion will be +1. Formula of the Li ion will be $\text{Li}^{+}$ .

Br is a group 17 nonmetal. As a nonmetal, it tends to form negative ions ("anions"). The charge on nonmetal ions excepting for H tends to equal the group number of the nonmetal minus 18. Br is in group 17. The charge on a Br ion will be 17 - 18 = -1. Formula of the Br ion will be $\text{Br}^{-}$

All the ions in an ionic compound carry charge. However, some of the ions like $\text{Li}^+$ are positive. Others ions like $\text{Br}^{-}$ are negative. Charge on the two types of ions balance each other. As a result, the compound is <em>overall</em> neutral.

1 × (+1) + 1 × (-1) = 0. The positive charge on one $\text{Li}^{+}$ ion balances the negative charge on one $\text{Br}^{-}$ ion. The two ions would pair up at a 1:1 ratio.

The empirical formula for an ionic compound shows all the ions in the compound. Positive ions are written in front of negative ions. $\text{Li}^{+}$ is positive and $\text{Br}^{-}$ is negative. The formula shall also show the simplest ratio between the ions. For the compound between Li and Br, a 1:1 ratio will be the simplest. The "1" subscript in an empirical formula can be omitted. Hence the formula: LiBr.

5 0

2 years ago

How many grams of ag can be formed from 5.50 grams of ag2o in the equation: 2ag2o (s) → 4ag (s) o2 (g)?

oee [108]

From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.

2Ag₂O(s) → 4Ag (s) + O₂ (g)

We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O= $\frac{5.50}{231.7}$ = 0.0237 moles.

From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces $\frac{4X0.0237}{2}$ moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.