How many moles of solute particles are present in 1 ml of aqueous 0.020 m (nh4)2co3?

1 answer:

6 0

Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?

For the purpose, here we will use the next equation:

c=n/V ⇒ n=cxV

n((NH4)2CO3) = 0.02M x 0.001L

n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution

You might be interested in

What type of spectrum is produced when electromagnetic radiation is emitted or absorbed as electrons change energy levels in an

Setler79 [48]

Then as the electrons in the atoms fall back down, they emit electromagnetic radiation (light). The amount of light emitted at different wavelengths, called the emission spectrum, is shown for a discharge tube filled with hydrogen gas in Figure 12.6 below.

8 0

2 years ago

Read 2 more answers

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$