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1 year ago

Which equation illustrates conservation of mass?

Chemistry

2 answers:

Flauer [41]1 year ago

7 0

Answer:

None as written.

Explanation:

Unless 2 meant H2 + Cl2 = 2HCl

pochemuha1 year ago

4 0

Answer = c

Conservation of mass (mass is never lost or gained in chemical reactions), during chemical reaction no particles are created or destroyed, the atoms are rearranged from the reactants to the products.

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How many moles of ions are in 285 ml of 0.0150 m mgcl2?

Solnce55 [7]

MgCl₂)= Mg²⁺ + 2Cl⁻
V(MgCl₂)=285cm³=0,285dm³
c(MgCl₂)=0,015 mol/dm³
n(MgCl₂)=c·V= 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol
n(Mg²⁺)=n(MgCl₂)=0,0042 mol
n(Cl⁻)=2n(MgCl₂)=0,0084 mol

7 0

2 years ago

Read 2 more answers

If a pharmacist dissolves 1.2 grams of a medicinal agent in 60 ml of a cough syrup having a specific gravity of 1.20, what is th

guapka [62]

Mass of medicinal agent taken = 1.2 g

the volume is 60 mL

Specific gravity = 1.20

So the mass of solution = specific gravity X volume = 1.20 * 60 = 72g

Now if we have increased the volume by 0.2 so the new volume = 60.2

New mass = 72 + 1.2 = 73.2

Specific gravity = mass / volume = 73.2 / 60.2 = 1.22 g/mL

7 0

2 years ago

The group of molecules called nucleotides contain:

Grace [21]

The group of molecules called nucleotides contain phosphate groups, pyrimidines, purines, and pentose (a 5-carbon sugar). Therefore, E. all of the above is the correct answer.

8 0

1 year ago

A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?

Rashid [163]

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

$\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}$

Given that:

number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent

volume of solution (HCl) = 450 mL 450 × 10⁻³ L

$\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}$

Normality N = 0.2 N

7 0

2 years ago

Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

nirvana33 [79]

Answer:

For 1: The correct answer is False.

For 2: The correct answer is True.

For 3: The correct answer is True.

For 4: The correct answer is False.

For 5: The correct answer is True.

Explanation:

Net ionic equation of any reaction does not include any spectator ions. If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

For 1: Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

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

Ionic form of the above equation follows:

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

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

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

Hence, the correct answer is False.

For 2: Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

$NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)$

Ionic form of the above equation follows:

$Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)$

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

For 3: Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

$Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)$

Ionic form of the above equation follows:

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

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

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

Hence, the correct answer is True.

For 4: Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

$MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

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

Hence, the correct answer is False.

For 5: Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

$(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the correct answer is True.

3 0

1 year ago