Answer:
[Ar] 3d10 4s2 4p3 is shorthand, 1s22s22p63s23p63d104s24p3 long hang
Explanation:
The shorthand is made using the lowest & closest noble gas, and picking up where it leaves off as follows, and longhand is made from a followed pattern you can easily find
Answer:

Explanation:
The breakdown reaction of ozone is as follows




It can be seen that 2 moles of ozone is required in the complete cycle
So for 10 cycles, 20 moles of ozone is required
m = Mass of
= 15.5 g
M = Molar mass of
= 104.46 g/mol
P = Pressure = 24.5 mmHg
T = Temperature = 232 K
R = Gas constant = 
Number of moles is given by


From ideal gas law we have

For 20 cycles of the reaction the volume of the ozone is
.
The molarity of the acid given the data from the question is 0.30 M
<h3>Balanced equation </h3>
2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HNO₃ (nA) = 2
- The mole ratio of the base, Ba(NO₃)₂ (nB) = 1
<h3>How to determine the molarity of the acid</h3>
From the question given above, the following data were obtained:
- Volume of acid, HNO₃ (Va) = 39.7 mL
- Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
- Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
- Molarity of acid, HNO₃ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 39.7) / (0.25 × 24) = 2
(Ma × 39.7) / 6 = 2
Cross multiply
Ma × 39.7 = 6 × 2
Ma × 39.7 = 12
Divide both side by 39.7
Ma = 12 / 39.7
Ma = 0.30 M
Learn more about titration:
brainly.com/question/14356286
#SPJ1
The statement is false because endothermic reactions have a positive enthalpy change, because heat is being added to the system because it requires more energy to react
Answer:
The new temperature will be 2546 K or 2273 °C
Explanation:
Step 1: Data given
The initial temperature = 1000 °C =1273 K
The volume = 20L
The volume increases to 40 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 20L
⇒with T1 = the initial temperature = 1273 K
⇒with V2 = the increased volume = 40L
⇒with T2 = the new temperature = TO BE DETERMINED
20L/ 1273 K = 40L / T2
T2 = 40L / (20L/1273K)
T2 = 2546 K
The new temperature will be 2546 K
This is 2546-273 = 2273 °C
Since the volume is doubled, the temperature is doubled as well