Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O
1) Ca-37, with a half-life of 181.1(10) ms.
Answer:
1.9 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 1.5 atm
- Initial volume (V₁): 3.0 L
- Initial temperature (T₁): 293 K
- Final pressure (P₂): 2.5 atm
- Final temperature (T₂): 303 K
Step 2: Calculate the final volume of the gas
If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂ / T₁ × P₂
V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L
Answer:
E
Explanation:
A catalyst is a substance which alters the rate of a chemical reaction. We check for the correctness of each of the options as follows:
A. Is wrong.
A catalyst can increase the rate of the forward and backward reaction
B. Is wrong
A catalyst does not slow the reverse reaction only. This particular case is the case of a negative catalyst
C is wrong
A catalyst has no effect on the equilibrium nor the equilibrium constant
D is wrong
Catalyst has no effect on equilibrium value
E is correct
Although Catalysts has no effect on equilibrium or its constant value, it can increase the the rate at which equilibrium is achieved by speeding up the reaction through bringing down the activation energy
