Answer:
Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}
or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4
Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.
Explanation:
Answer:
209.3 Joules require to raise the temperature from 10 °C to 15 °C.
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass of water = 10 g
initial temperature T1= 10 °C
final temperature T2= 15 °C
temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C
Energy or joules added to increase the temperature Q = ?
Solution:
We know that specific heat of water is 4.186 J/g .°C
Q = m × c × ΔT
Q = 10 g × 4.186 J/g .°C × 5 °C
Q = 209.3 J
The shape of BeH2 is linear.
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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