19E molar/molal Test

How many grams of agcl would be needed to make a 4. 0 m solution with a volume of 0. 75 l?

devlian [24]

430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

<h3>What is Molarity?</h3>

The amount of a substance in a specific volume of solution is known as its molarity (M).

The number of moles of a solute per liter of a solution is known as molarity.

<h3>Calculation of Required amount of AgCl</h3>

Remember that mol/L is the unit of molarity (M).

We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.

0.75L×4.0M=3.0mol

Then, using the periodic table's molar mass for AgCl, convert from moles to grams:

3.0mol×143.321gmol=429.963g

The final step is to round to the correct significant figure, which in this case is two: 430g.

Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

Learn more about Molarity here:
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8 0

1 year ago

BARINLY PLS HELP ME ASP !!! it’s science I need help fast pls

Oliga [24]

Answer:

Its D. All of the above

8 0

3 years ago

What could using cobalt glass be helpful to identify?<br>

satela [25.4K]

Answer:

Using cobalt glass could be helpful to identify elements that weakly emit blue and/or violet.

Explanation:

5 0

2 years ago

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins wi

Marizza181 [45]

Answer: 3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced. Explanation: Ostwald process is a multi-step for manufacturing Nitric Acid. First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow. Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) ⟶ NO2(g) Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g) For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms. So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides. Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l) Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g) Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g) Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg and Molecular Weigth = 63,01g/mol NH3 Molecular Weight= 17,031 g/mol Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights. 7,839Kg of HNO3 x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3 The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3

7 0

3 years ago

Write the equation for the dissociation of hcl when hcl is dissolved in water.

kotykmax [81]

Answer:

HCl -> H+ + Cl- (monoprotic acid)

3 0

2 years ago