Answer:
2NaOH + CO2 -> Na2CO3 + H2O
1) Find the moles of each substance
2) Determine the limitting reagent
∴ Carbon dioxide is limitting as it has a smaller value.
3) multiply the limiting reagent by the mole ratio of unknown over known
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
4) Multiply the number of moles by the molar mass of the substance.
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g
Explanation:
Answer:
0.0295M
Explanation:
As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.
You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.
That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.
And the concentration of SCN⁻ is:
0.0800M / 2.714 =
<h3>0.0295M </h3>
Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:
Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M
Answer:
±0.005 g
Explanation:
The uncertainty depends on whether the measurement was obtained manually or digitally.
1. Manual
The minimum uncertainty is ±0.01 g.
It may be greater, depending on random or personal errors
2. Digital
Most measurements of mass are now made on digital scales.
A digital device must always round off the measurement it displays.
For example, if the display reads 20.00, the measurement must be between 20.005 and 19.995 (±0.005).
If the measured value were 20.006, the display would round up to 20.01.
If the measured value were 19.994, the display would round down to 19.99.
The uncertainty is ±0.005 g.
The scale shown below would display a mass of 20.00 g
Hey there!
Values Ka1 and Ka2 :
Ka1 => 8.0*10⁻⁵
Ka2 => 1.6*10⁻¹²
H2A + H2O -------> H3O⁺ + HA⁻
Ka2 is very less so I am not considering that dissociation.
Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]
lets concentration of H3O⁺ = X then above equation will be
8.0*10−5 = [x] [x] / [0.28 -x
8.0*10−5 = x² / [0.28 -x ]
x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵
solve the quardratic equation
X =0.004693 M
pH = -log[H⁺]
pH = - log [ 0.004693 ]
pH = 2.3285
Hope that helps!