Balanced equation : C. CH₄ + 4Cl₂⇒ CCl₄+ 4HCl
<h3>Further explanation </h3>
Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:
1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.
2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product
3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
CH₄ + Cl₂⇒ CCl₄+ HCl
aCH₄ + bCl₂⇒ CCl₄+ cHCl
C, left=a, right=1⇒a=1
H, left=4a, right=c⇒4a=c⇒4.1=c⇒c=4
Cl, left=2b, right=4+c⇒2b=4+c⇒2b=4+4⇒2b=8⇒b=4
The equation becomes :
CH₄ + 4Cl₂⇒ CCl₄+ 4HCl
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>
Explanation:
The given data is as follows.
Concentration of solution = 0.5 M
Volume of solution = 1 L
Molar mass of Glycylglycine = 132.119 g/mol
As molarity is the number of moles present in liter of solvent.
Mathematically, Molarity = 
Hence, calculate the number of moles as follows.
No. of moles = Molarity × Volume
= 
= 0.5 mol
Therefore, mass of glycylglycine = mol × molar mass
= 
= 66.06 g
Thus, we can conclude that 66.06 g glycylglycine is required.
Mass = m = 40 grams
Volume = V = 9 cm³
Density = d = ?
Density is defined as the ratio of mass and volume.
So,
d = m/V
Using the values, we get
d = 40/9 = 4.44 g/cm³
This means the density of material would be 4.44 g/cm³
