Answer:
0.01144L or 1.144x10^-2L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 20.352 mL
P1 (initial pressure) = 680mmHg
P2 (final pressure) = 1210mmHg
V2 (final volume) =.?
Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:
P1V1 = P2V2
680 x 20.352 = 1210 x V2
Divide both side by 1210
V2 = (680 x 20.352)/1210
V2 = 11.44mL
Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:
1000mL = 1 L
11.44mL = 11.44/1000 = 0.01144L
Therefore the volume of the container is 0.01144L or 1.144x10^-2L
Answer:
Ag
Explanation:
To determine which element will displace hydrogen from a dilute acid, we need to make reference to the activity series or the electrochemical series. The activity series is a list of metallic ions according to their electropositivity.
This means elements are ranked here based on how electrically positive they are. Hence, an element above another element is relatively more electropositive than the one under it. For example calcium is less electropositive compared to sodium as calcium is found under sodium in the list.
Now, for an element to displace hydrogen, it means the particular element is more electropositive than hydrogen on the activity series. All the elements in the options are in a greater position relative to hydrogen on the activity series except silver. This means it cannot displace hydrogen from a dilute mineral acid
Explanation:
your answer is Kelvin because it is the SI unit of temperature
Enthalpy change during the dissolution process = m c ΔT,
here, m = total mass = 475 + 125 = 600 g
c = <span>specific heat of water = 4.18 J/g °C
</span>ΔT = 7.8 - 24 = -16.2 oc (negative sign indicates that temp. has decreases)
<span>
Therefore, </span>Enthalpy change during the dissolution = 600 x 4.18 X (-16.2)
= -40630 kJ
(Negative sign indicates that process is endothermic in nature i.e. heat is taken by the system)
Thus, <span>enthalpy of dissolving of the ammonium nitrate is -40630 J/g</span>
