Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.
The azide ion that is 
, is a symmetrical ion, all of whose contributing structures have formal charges.
Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.
Contributing structures of azide ion are drawn in the image attached.
Answer:
92.6
Explanation:
6 mol x 18.02 g of H2o --> 3 mol x 58.33 g Mg(OH)2
108.12 g of h2o --> 174.99 of Mg(OH)2
g of H2O is 150 g of Mg(OH)2
150g x 108.12g / 174.99 =
92.67
The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.
Answer:
4.48 - 6.48
Explanation:
A pH indicator works in a better way in a range of pH = pKa ± 1. That means we need to determine the pKa of the indicator propyl red to find the range over which it change its color. That is:
pKa = -log Ka
pKa = -log 3.3x10⁻⁶
pKa = 5.48
That means the range over propyl red will change from yellow to red or vice versa is:
4.48 - 6.48
