Calculate the mass of mgco3 precipitated by mixing 10.00 ml of a 0.200 m na2co3 solution with 5.00 ml of a 0.0450 m mg(no3)2 solution.

1.90 g

Explanation:

When we are making calculations based on chemical reactions, we need first to have a balanced chemical equation.

From there we can determine the mass of the product MgCO₃ in this question.

Na₂CO₃   +   Mg(NO₃)₂     ⇒   2 NaNO₃   + MgCO₃ ( double decomposition )

0.200 M   0.0450 M                                        ?

10.0           5.00 mL  

Now we know the volume and concentration of Mg(NO₃)₂,and Na₂CO₃ so we must compute their number of moles to determine the limiting reagent, if any.From there determine moles and mass of MgCO₃ produced.

First lets convert the volume of  Mg(NO₃)₂and Na₂CO₃ to liters:

5.00 mL x ( 1 L/1000 mL ) =    5.00 x 10⁻³ L

10.00 mL x ( 1L/ 1000 mL ) = 1.000 x 10 ⁻² L

# mol Mg(NO₃)₂ = ( 0.0450 mol /L  ) x 5.00 x 10⁻³ L

                                         = 2.25 x 10⁻⁴ mol Mg(NO₃)₂

# mol Na₂CO₃ = ( 0.200 mol / L ) x 1 x 10⁻² L  

                                         = 2.000 x 10⁻³  mol Na₂CO₃

Calculation limiting reagent:

= 2.25 x 10⁻⁴ mol Mg(NO₃)₂ x ( 1 mol Na₂CO₃ / mol  Mg(NO₃)₂ )

= 2.25 x 10⁻⁴ mol  Na₂CO₃ required to react

Therefore, our limiting reagent is Mg(NO₃)₂ since we require  2.25 x 10⁻⁴ mol  Na₂CO₃  to react completely with  2.25 x 10⁻⁴Mg(NO₃)₂, and we have excess of it.

# mol MgCO₃ produced

= 2.25 x 10⁻⁴ mol  Mg(NO₃)₂ x ( 1 mol MgCO₃ / 1 mol Mg(NO₃)₂ )

= 2.25 x 10⁻⁴ mol  MgCO₃

Now that we have the moles of MgCO₃, we can obtain its mass by multiplying its molar mas ( 84.31 g/mol ):

2.25 x 10⁻⁴ mol  MgCO₃ x   84.31 g/mol  = 1.90 g

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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