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atroni [7]
3 years ago
7

A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita

te. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) How many grams of precipitate potentially form?
Chemistry
1 answer:
mestny [16]3 years ago
4 0

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

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Answer:

V₂ = V₁ / T₁ * T₂ . If you prefer to set the final volume and want to estimate the resulting temperature, then the equation of Charles' law changes to: T2=T1/T1 multiplied by v^2.

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1 year ago
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In the molecular orbital model of cyclobutadiene, how many -antibonding molecular orbitals are there?
Citrus2011 [14]

There are one antibonding molecular orbitals present in molecular orbital model of c.

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4 0
2 years ago
The 85.2-g sample of the compound x4o10 contains 48.0 g of oxygen atoms. what is the molar mass of element x?
Vedmedyk [2.9K]
X4O10
Let molar mass of X be y
molar mass =  4y + 10 x 16 = 4y+160

so, moles = 85.2 / (4y+160)

Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen  = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0

so, 48 = 16 x 10 x [85.2 / (4y+160) ]

Solve the equation to get y.

y = 31
5 0
2 years ago
We can change a gas to liquid by_________ the temperature and_________
Tpy6a [65]
Decreasing, Removing
6 0
3 years ago
The equilibrium constant for the reaction
FinnZ [79.3K]

Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

Moles of  Cl_2 = \frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol

Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

Now put all the given values in this expression, we get :

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

7 0
3 years ago
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