Answer:
The final temperature is 20.3 °C
Explanation:
Considering that:-
Heat gain by water = Heat lost by brass
Thus,
Where, negative sign signifies heat loss
Or,
For water:
Mass = 300.0 g
Initial temperature = 20.0 °C
Specific heat of water = 4.184 J/g°C
For brass:
Mass = 19.0 g
Initial temperature = 81.7 °C
Specific heat of water = 0.375 J/g°C
So,
<u>Hence, the final temperature is 20.3 °C</u>
Above question is incomplete. Complete question is attached below
........................................................................................................................
Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case,
reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is
least reactive.
On other hand,
reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is
most reactive.
Answer:
![r=-k[H_2][O_2]^{1/2}](https://tex.z-dn.net/?f=r%3D-k%5BH_2%5D%5BO_2%5D%5E%7B1%2F2%7D)
Explanation:
Hello!
In this case, according to the chemical reaction:
H2 + 1/2 O2 --> H2O
Since both hydrogen and oxygen are the reactants and the rate laws are written in terms of the concentration of reactants and the rate constant, we obtain:
Whereas the negative sign represents the consumption of those reactants and the power 1/2 the stoichiometric coefficient of oxygen in the reaction.
Best regards!
The mass of 63 ml sample : 79.38 g
<h3>Further explanation</h3>
Given
20 ml and 25.2 g of glycerol
Required
The mass of 63 ml sample
Solution
Density is the ratio of mass per unit volume
Density formula:
Density of glycerol :
= m : V
= 25.2 g : 20 ml
= 1.26 g/ml
Mass of 63 ml sample :
= density x volume
= 1.26 g/ml x 63 ml
= 79.38 g
Answer:
B. 0.5 molar
Explanation:
Given data:
Initial concentration = 0.40 M
Initial volume = 750 mL
Final volume =750 - 150 mL = 600 mL
Final concentration = ?
Solution:
Molarity is the number of moles of solutes in litter of solvent. In given problem it is stated that when the solution is uncovered solvent evaporate it means molarity is changed. we can calculate the new molarity with the following formula.
C₁V₁ = C₂V₂
C₁ = initial concentration
V₁ = initial volume
C₂ = final concentration
V₂ = final volume
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.40 M × 750 mL = C₂ × 600 mL
300 M.mL / 600 mL = C₂
0.5 M = C₂
