Answer:
TABLE OF CONJUGATE ACID-BASE PAIRS Acid BaseK a (25oC) HClO 4ClO 4 – H 2 SO 4HSO 4 – HCl Cl– HNO 3NO 3 – H 3 O +H 2 O H 2 CrO 4HCrO 4 –1.8 x 10–1
Explanation:
When 400 J of heat are slowly added to 10 mol of an ideal monatomic gas, its temperature rises by 10°C. The work done on the gas is 845J.
The amount of work done on gas depends upon the internal energy change and the heat supplied to the system.
According to First Law of Thermodynamics, the change in internal energy is equal to the work done and the heat supplied to the system.
This is given by:
ΔU = W + Q
where, ΔU is change in Internal energy
W is the Work done
Q is the heat supplied
Given,
Q = 400J
Number of moles, n = 10
Change in temperature, ΔT = 10°C
Cv = 3/2R ; Since, the given gas is monoatomic (R=8.3)
We know that, ΔU = n Cv ΔT
On substituting the values in above formula,
ΔU = 10 × 3/2 × 8.3 × 10
ΔU = 1245J
Using,
ΔU = W + Q
1245J = W + 400
W = 845J
Hence, the work done on the gas is 845J.
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H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
Explanation:
Unbalanced reaction equation:
H₂SO₄ + NH₄OH → (NH₄)₂SO₄ + H₂O
We can use a simple mathematical approach to balance this equation by solving simple algebraic equations.
aH₂SO₄ + bNH₄OH → c(NH₄)₂SO₄ + dH₂O
a,b,c and d are simple coefficients that will balance the equation:
Conservation of H: 2a + 5b = 8c + 2d
S: a = c
O: 4a + b = 4c + d
N: b = 2c
lets assume that a = 1
c= 1
b = 2
Solving for d from 4a + b = 4c + d
d = 4a + b -4c = 4(1) + 2 - 4(1) = 2
H₂SO₄ + 2NH₄OH → (NH₄)₂SO₄ + 2H₂O
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Answer: solvent will move from the first compartment containing 3% glucose into the second compartment containing 6% glucose.
Explanation: osmosis is defined as the movement of solvent (example water molecules) from a region of lower concentrated solution into the region of more concentrated solution. The first compartment contains less glucose concentration with more solvent. Therefore, solvent will move from the first compartment with less glucose concentration into the second compartment containing 6% glucose concentration to achieve an equalised concentration of solute on both sides of the semi permeable membrane.
