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3 years ago

If the density of an object is 5.2g/cm3, and it’s volume is 3.7 cm3, what is it’s mass?

1 answer:

5 0

Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3

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2. How many moles of salt are present in 1.5L of a 5.OM salt water solution?

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Answer:

There are 7.5 moles of salt

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3 years ago

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

mestny [16]

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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