The mass of iron block is 500 g. The amount of energy required to melt the iron block needs to be calculated. Melting means conversion of solid to liquid thus, heat of fusion is used which is 247 J/g.
From heat of fusion, 247 J of energy is released by melting 1 g of iron block. Thus, the amount of heat released by melting 500 g of iron rod will be:
H= 247 J/g× 500 g=1.23×10^{5}
Hence, option B is correct.
The chemical formula for the compound can be written as,
CxHyOz
where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
CxHyOz + O2 --> CO2 + H2O
number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C
number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O
This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H
Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g
Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O
The formula of the compound is,
C0.0163H0.01635O0.0109
Dividing the numbers by the least number,
C3/2H3/2O
The empirical formula of the compound is therefore,
<em> C₃H₃O₂</em>
PH is calculated using <span>Handerson- Hasselbalch equation,
pH = pKa + log [conjugate base] / [acid]
Conjugate Base = Acetate (CH</span>₃COO⁻)
Acid = Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]
We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,
pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵
pKa = 4.75
Now,
Putting all values in eq. 1,
pH = 4.75 + log [0.172] / [0.818]
pH = 4.072
Answer:
No
Explanation:
If tin is heated, it can react with alkalis' with the release of hydrogen.
Answer: 1.14
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
To calculate pH of gastric juice:
molarity of 
= 0.072
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Thus the pH of the gastric juice is 1.14
