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Stells [14]
3 years ago
5

Which of the following has the highest concentration?

Chemistry
1 answer:
Andreyy893 years ago
5 0

Answer:

ℂ

Explanation:

ℂ

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after a chemistry student made a AgNO3(small 3 at bottom) solution., she wanted to determine the molar concentration of it. if 2
Alexxandr [17]
Molar concentration = (numbet of mol Solute)/ ( volume Solution)

1) Finding
 the number of the mol solute

22.0 g AgNO3 * \frac{1 mol AgNO3}{169.91 g AgNO3} = 0.129 mol AgNO3

2) 725 ml = 0.725 L




3) Molarity = \frac{number/ mol/solute}{Volume/solution} = \frac{0.129}{0.725} 



Molarity= \frac{0.178 mol}{L} , 

or 0.178 M



3 0
4 years ago
Which elements are<br> considered "Noble Metals"?
Lisa [10]

Answer:

ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).

Explanation:

7 0
3 years ago
Revolver features several firing chambers located within...
Blizzard [7]

Answer:

What are the characteristics of a revolver?

Explanation:

The revolver features several firing chambers located within a revolving cylinder. The cylinder rotates as the revolver is fired, lining up each bullet mechanically.

8 0
3 years ago
The reaction between A2 (two large red spheres) and B2 (two small blue spheres) to produce AB is shown in the diagram. The react
drek231 [11]

Answer:

B₂

Explanation:

The limiting reactant is always a reactant. You can determine which reactant is limiting by identifying which has the smaller mole-to-mole ratio with the product. This ratio can be found via the coefficients of the balanced reaction.

4 A₂ + 3 B₂ ---> 6 AB

4 moles A₂
------------------  = mole-to-mole ratio A₂/AB
6 moles AB

3 moles B₂
------------------  =  mole-to-mole ratio B₂/AB
6 moles AB

Since the mole-to-mole ratio between B₂ and AB is smaller, B₂ must be the limiting reactant.
                           

3 0
2 years ago
Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a
Crank

Explanation:

a. Pb(NO_3)_2(aq) + Na_2SO_4(aq) → ?

Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)

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

Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)

Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)

b. NiCl_2(aq) + NH_4NO_3(aq) →

NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)

No precipitation is occuring.

c. Fe_Cl2(aq) + Na_2S(aq) →

FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)

FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)

Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)

Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)

d.MgSO_4(aq) + BaCl_2(aq) →

MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2

MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)

BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)

Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)

Removing common ions from both sides, we get the net ionic equation:

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

5 0
3 years ago
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