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3 years ago

45 grams of ammonia, NH3, in 0.75 L of solution.

1 answer:

4 0

2.3 moles of sodium chloride in 0.45 liters of solution . necessarile correct ... 0.025 L. (1.3 SF). 45 grams of ammonia in 0.75 L of solution. 45g x 1 mol = 2. demol ... anol x 1,50 = 3 mol NaOH x 40,019 = 120g Naolt.

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3 years ago

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3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

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n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

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n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

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2 years ago

I need help, I don't understand the question

Oduvanchick [21]

Remember this.

Ionic molecules has ionic bonds
Nonpolar molecules has dispersion (Van del Waals)
Polar molecules could either have hydrogen bonding or Dipole-Dipole. Hydrogen bonding is when you have F, O or N with H, every other polar molecule is dipole-dipole.

a. polar- dipole-dipole
b. polar- hydrogen bonding
c. nonpolar- dispersion
d. nonpolar- dispersion
e. polar- dipole-dipole
f. polar-dipole-dipole
g. nonpolar- dispersion
h. polar- hydrogen bonding.

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3 years ago