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2 years ago

Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution?

2 answers:

8 0

Mass = molarity x molar mass( NaNO₃) x volume

mass = 1.50 x 85.00 x 4.50

mass = 573.75 g of NaNO₃

hope this helps!

Fynjy0 [20]2 years ago

4 0

Answer:

D.) 547 g

Explanation:

this is the right answer on edg 2020, just took quiz

hope this helps!!!

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A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= $\frac{4.00 g}{245 g/mol}=0.01633 mol$

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

$Molarity=\frac{n}{V(L)}$

Let the molarity of the solution be $M_1$

$M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M$

A students then takes 4 .00 mL of $M_1$ solution and dilute it to 275 ml.

$M_1=0.1633 M$

$V_1=4.00 mL$

$M_2=?$ (molarity after dilution)

$V_2=275 mL$ (after dilution)

$M_1V1=M-2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M$

Molarity of the of solution after dilution is 0.002375 M.

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

$[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M$

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = $[NO_3^{-}]=0.004750 M$

$[NO_3^{-}]=\frac{n}{0.275 L}$

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097 grams of nitrate ions are there in the final solution.

4 0

2 years ago

How many chloride ions are in 0.486 moles of chloride ions?

svetlana [45]

Answer:

Since in a chloride ion, we have an additional electron

you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that

Amount of ions in 1 mol = 6.022 * 10^23

Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)

Amunt of ions in 0.486 moles = 2.9 * 10^23 ions

Hence, option 1 is correct

6 0

2 years ago

How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?

svet-max [94.6K]

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)

assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3, are the minimun moles necessary to neutralize the acid.

6 0

2 years ago

To construct the galvanic cell illustrated above, the salt bridge was prepared by soaking a piece of cotton in 5.0MNaNO3(aq) bef

dalvyx [7]

Answer:

The cell reaction reaches equilibrium quickly and the cell emf becomes zero.

Explanation:

The purpose of a salt bridge is not to move electrons from the electrolyte, its main function is to maintain charge balance because the electrons are moving from one-half cell to the other.

A solution of a salt that dissociates easily is normally used. Water is ineffective at functioning as a salt bridge. Hence the effect stated in the answer.

4 0

2 years ago

Monochlorination of propane yields two constitutional isomers, and dichlorination yields four. Trichlorination yields five const

GaryK [48]

By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.

6 0

2 years ago

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