Lithium has charge of +1 and bromide has charge of - 1. So they combine to form the compound lithium bromide which is expressed as LiBr.
<h3><u>Explanation:</u></h3>
Lithium is an alkali metal placed in group 1 or periodic table. It has a valency of 1 which is achieved as lithium loses an electron to achieve a charge of +1.
Bromine is a halogen which is placed in group 17 of periodic table. It has a valency of 1 which is achieved as bromine looses an election to achieve a charge of - 1.
Lithium is the cation and bromide is the anion. So lithium is written in front and bromine following the cation. And as both of their valencies are 1, so they form the compound LiBr.
Answer:
The answer to your question is P2 = 1.52 atm
Explanation:
Data
Volume 1 = V1 = 4.54 l
Pressure 1 = P1 = 1.65 atm
Temperature 1 = T1 = 75°C
Volume 2= V2 = 5.33 l
Pressure 2 = P2 = ?
Temperature 2 = 103°C
Process
1.- Convert temperature to °K
Temperature 1 = 75 + 273 = 348°K
Temperature 2 = 103 + 273 = 376°K
2.- Use the combine gas law to find the final pressure
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (1.65 x 4.54 x 376) / (348 x 5.33)
-Simplification
P2 = 2816.62 / 1854.84
-Result
P2 = 1.52 atm
Answer:
pH = 1.683
Explanation:
plz mark as brainliest if it helps
Answer:
1023.75mmHg
Explanation:
V1 = 3.5L
P1 = 585mmHg
V2 = 2.0L
P2 = ?
To solve this question, we'll require the use of Boyle's law which states that the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature is kept constant.
Mathematically,
V = kP, k = PV
P1 × V1 = P2 × V2 = P3 × V3 = .......= Pn × Vn
P1 × V1 = P2 × V2
Solve for P2,
P2 = (P1 × V1) / V2
P2 = (585 × 3.5) / 2.0
P2 = 2047.5 / 2.0
P2 = 1023.75mmHg
The final pressure of the gas is 1023.75mmHg
Answer:
The assumption is quite reasonable.........
A lightbulb contains Ar gas at a temperature of 295K and at a pressure of 75kPa. The light bulb is switched on, and after 30 minutes its temperature is 418 K. What is a numerical setup for calculating the pressure of the gas inside the light bulb at 418K?
Explanation:
P
1
T
1
=
P
2
T
2
given constant
n
, and constant
V
, conditions that certainly obtain with a fixed volume light bulb.
And so
P
2
=
P
1
T
1
×
T
2
=
75
⋅
k
P
a
295
⋅
K
×
418
⋅
K
≅
100
⋅
k
P
a
.
Had the light bulb been sealed at normal pressure during its manufacture, what do you think might occur when it is operated?
