Answer:
Option D. 30 mL.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above,
The mole ratio of the acid, nA = 1
The mole ratio of the base, nB = 1
Step 2:
Data obtained from the question. This include the following:
Volume of base, KOH (Vb) =.?
Molarity of base, KOH (Mb) = 0.5M
Volume of acid, HNO3 (Va) = 10mL
Molarity of acid, HNO3 (Ma) = 1.5M
Step 3:
Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:
MaVa / MbVb = nA/nB
1.5 x 10 / 0.5 x Vb = 1
Cross multiply
0.5 x Vb = 1.5 x 10
Divide both side by 0.5
Vb = (1.5 x 10) /0.5
Vb = 30mL
Therefore, the volume of the base, KOH needed for the reaction is 30mL.
Answer:
% I = 0.083 %
Explanation:
- C6H5COOH + NaOH ↔ NaC6H5CO2 + H2O
∴ M C6H5COOH = 0.18 mol/L
∴ M NaC6H5CO2 = 0.10 mo/L
- % ion = ( { H3O+ ] / initial acid concentration ) * 100
⇒ C6H5COOH ↔ H3O+ + C6H5COO-
⇒ NaC6H5CO2 ↔ Na+ + C6H5COO-
∴ Ka = 6.4 E-5 = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ]
Ka value taken from the literature
mass balance
⇒ 0.18 + 0.1 = [ C6H5COOH ] + [ C6H5COO- ]
⇒ [ C6H5COOH ] = 0.28 - [ C6H5COO- ] ........(1)
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ C6H5COO- ]
∴ [ Na+ ] ≅ M NaC6H5CO2 = 0.1 M
⇒ [ C6H5COO- ] = [ H3O+ ] + 0.1..............(2)
(1) and (2) in Ka:
⇒ 6.4 E-5 = ( [ H3O+ ] * ( [ H3O+ ] + 0.1 ) / ( 0.28 - ( [ H3O+ ] + 0.1 ) )
⇒ 6.4 E-5 = [ H3O+ ]² + 0.1 [H3O+ ] / ( 0.18 - [ H3O+ ] )
⇒ 1.17 E-5 - 6.5 E-5 [ H3O+ ] = [ H3O+ ]² + 0.1 [ H3O+ ]
⇒ [ H3O+ ]² + 0.1 [ H3O+ ] - 1.17 E-5 = 0
⇒ [ H3O+ ] = 1.493 E-4 M
⇒ % I = ( 1.493 E-4 / 0.18 ) * 100 = 0.083 %
Answer:
Tbh i rlly dont think that many will. Most may succeed in doing so. But i mean come on its 2020 anything can happen.
Explanation:
Answer:
Because the specific heat of the metal is less than the specific heat of water.
Explanation:
Hello, happy to help you today!
In this case, we need to analyze a property called "specific heat" which accounts for how much energy is required to increase or decrease the temperature of 1 g of the substance by 1 °C.
In this case, since the specific heat of water is about 4.184 J/g°C and the specific heat of metals in general is greater than zero, of course, but less than one, we can infer that for the same amount of energy, when they are in contact, more grams of metal will be cooled down to those of water heated up, because the specific heat of the metal is less than the specific heat of water.
Best regards!.
Detects? i think so but don't quote me.
