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3 years ago

Find the density of an object that has a mass of 28 grams and a volume of 2 points 38.6 cm^3. *

Chemistry

1 answer:

Shtirlitz [24]3 years ago

4 0

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{28}{38.6} \\ =0.72538...$

We have the final answer as

Hope this helps you

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Does the volume of a liquid change

nydimaria [60]

Answer:

Yes it does change

Explanation:

6 0

4 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

<u>Answer:</u>

<u>For (a):</u> The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

<u>For (b):</u> The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

<u>For (c):</u> The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

<u>For (d):</u> The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

<u>For (e):</u> The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

<u>Explanation:</u>

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago

When 8.0 grams of sodium hydroxide is dissolved in sufficient water to make 400. mL of solution, what is the concentration of th

nata0808 [166]

Answer:

The concentration of this sodiumhydroxide solutions is 0.50 M

Explanation:

Step 1: Data given

Mass of sodium hydroxide (NaOh) = 8.0 grams

Molar mass of sodium hydroxide = 40.0 g/mol

Volume water = 400 mL = 0.400 L

Step 2: Calculate moles NaOH

Moles NaOH = mass NaOH / molar mass NaOH

Moles NaOH = 8.0 grams / 40.0 g/mol

Moles NaOh = 0.20 moles

Step 3: Calculate concentration of the solution

Concentration solution = moles NaOH / volume water

Concentration solution = 0.20 moles / 0.400 L

Concentration solution = 0.50 M

The concentration of this sodiumhydroxide solutions is 0.50 M

6 0

3 years ago

When was the idea of a atom first devloped

bija089 [108]

Answer:

Around 450 B.C.

Explanation:

The idea was forgotten until the 1800 when John Dalton re-introduced the atom.

6 0

3 years ago

Identify the name of the compound Cr2(CO3)3.

kompoz [17]

Cr2(CO3)3 is Chromium (III) Carbonate or Chromic Carbonate.

It is composed of Chromium, Carbon, and Oxygen. Chromium has 2 atoms, Carbon has 3 atoms, and Oxygen has 9 atoms.

This compound is composed of 36.6% Chromium, 12.7% Carbon, and 50.7% Oxygen.

7 0

4 years ago

Read 2 more answers