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4 years ago

How many grams of sodium chloride are present in a 0.75 M solution with a volume of 500.0 milliliters?

1 answer:

6 0

The molarity of a solution is a type of expression of concentration equal to the number of moles solute per liter solution. In this problem, we are given the molarity equal to 0.75 M and a volume equal to 500 milliliters. <span>500 milliliters is equal to 0.5 liters. we multiply M and L to get the number of moles then multiply by the molar mass of NaCl. The answer is 21.92 grams.</span>

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Examine the reaction. NH4OH(aq) →H2O(l) + NH3(g)

shusha [124]

Answer: A. 1,1,1

Explanation:

The coefficients that will balance the equation; NH4OH(aq) →H2O(l) + NH3(g), is 1, 1, 1, because it proves the total number of atoms of each element on the LHS and RHS of the equation are equal, hence balanced.

LHS RHS

N = 1 1

H = 5 5

O = 1 1

7 0

3 years ago

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago

Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:

Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

$\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}$

$\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21$

8 0

3 years ago

Describe three types of POTENTIAL ENERGY

Arada [10]

There are three different forms of potential energy. The rock hanging above the ground has a form of stored energy called gravitational potential energy. This form of energy is due to the downward pull of Earth's gravity. ... When you stretch a rubber band, the elastic potential energy of the rubber band increases.

8 0

3 years ago

In a series circuit there are 17 light bulbs, each providing 1 ohm of resistance. What is the total resistance?

Sauron [17]

Answer:

17 ohms

Explanation:

Given that,

In a series circuit there are 17 light bulbs, each providing 1 ohm of resistance.

We need to find the total resistance of the circuit.

The equivalent resistance for series combination is given by :

$R_{eq}=R_1+R_2+....$

For 17 light bulbs, the equivalent resistance is given by :

$R_{eq}=R+R+....\text{upto 17 terms}\\\\R_{eq}=17R\\\\=17$

Hence, the total resistance is 17 ohms. Hence, the correct option is (d).

7 0

3 years ago