Answer:
false
Explanation:
not too sure sorry if im wrong
Answer:
THE MOLARITY IS 2.22 MOL/DM3
Explanation:
The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water
Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol
Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;
37.5 g of Na2S = 217 g of water
( 37.5 * 1000 / 217 ) g = 1000 g of water
So, 172.81 g/dm3 of the solution
So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass
Molarity = 172.81 g/dm3 / 78 g/mol
Molarity = 2.22 mol/dm3
The molarity of the solution is 2.22 mol/dm3
Answer:
625.46 °C
Explanation:
We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 19 °C
T(K) = 19 °C + 273
T(K) = 292 K
Next, we shall determine the Final temperature. This can be obtained as follow:
Initial volume (V₁) = 3.25 L
Initial temperature (T₁) = 292 K
Final volume (V₂) = 10 L
Final temperature (T₂) =?
V₁/T₁ = V₂/T₂
3.25 / 292 = 10 / T₂
Cross multiply
3.25 × T₂ = 292 × 10
3.25 × T₂ = 2920
Divide both side by 3.25
T₂ = 2920 / 3.25
T₂ = 898.46 K
Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 898.46 K
T(°C) = 898.46 – 273
T(°C) = 625.46 °C
Therefore the final temperature of the gas is 625.46 °C
Atomic mass number is the number of protons and neutrons. Subtract 80-35=45 is the number of protons. Because the atom is neutrally charged, the number of protons must equal the number of electrons, so there are 45 electrons.
Answer: -
The rate decreases as the concentration of the reactants decreases
Explanation: -
A reaction involves change of the reactants into products.
Initially there is only reactants. So the rate if reaction is high.
After some time there are products. So the amount of reactant is less.
Reactions involve collisions of reactant molecules. As the reactant amount decreases, collisions between the reactants decreases. As such the rate of reaction decreases with the progress of the reaction.
