Answer:
A chemical equation is balanced when the number of each kind of atom is the same on both sides of the reaction.
Explanation:
The law of conservation of matter (except in nuclear reactions) indicates that atoms can neither be created or destroyed.
The number of atoms that are in the reactants must be the same as the number of the atoms that are in the product.
The number and types of molecules can (and will) change. The atoms that make up the molecules are rearranged but the number and kinds of atoms stay the same.
Answer:
K, the rate constant = 9.73 × 10^(-1)/s
Explanation:
r = K × [A]^x × [B]^y
r = Rate = 1.07 × 10^(-1)/s
K = Rate constant
A and B = Concentration in mol/dm^-3
A = 0.44M
B = 0.11M
x = Order of reaction with respect to A = 0
y = Order of reaction with respect to B = 1
Solving, we get
r/([A]^x × [B]^y) = K
K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727
K = 0.9727
Answer:
Explanation:B ice melting is a phiyscal change and the mass of the substance remains the same
A) Energy is released during the formation of the bond.
Explanation:
During the formation of a chemical bonds between two hydrogen atoms, energy is always released during the formation of this bond type.
Bond formation process is usually exothermic and energy is released during the formation of the bond.
- Bond breaking process is an endothermic process in which energy is absorbed from the surrounding.
- Whenever a bond is broken, the bond energy value is positive but when a bond is formed, the bond energy value is given a negative sign.
For a bond formation process in which hydrogen atoms are bonded covalently, energy is usually released.
Learn more:
Enthalpy changes brainly.com/question/10567109
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Answer:
pH= 11.49
Explanation:
Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.
From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5
Using the formula below;
[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)
[OH^-] =√(3.2×10^-5 × 0.30M)
[OH^-]= √(9.6×10^-6)
[OH^-]=3.0984×10^-3
pOH= -log[OH^-]
pOH= -log 3.1×10^-3
pOH= 3-log 3.1
pH= 14-pOH
pH= 14-(3-log3.1)
pH= 11+log 3.1
pH= 11+ 0.4914
pH= 11.49
