To solve this we assume that the hydrogen gas is an
ideal gas. Then, we can use the ideal gas equation which is expressed as PV =
nRT. At a constant pressure and number of moles of the gas the ratio T/V is
equal to some constant. At another set of condition of temperature, the
constant is still the same. Calculations are as follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K
<span>V2 = 12.09 L</span>
Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.
<span />
Answer:
V1= 0.305L
Explanation:
To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2
Since the question wants the volume in litres, convert 455 mL to L
455/ 1000
= 0.455 L
Now make the substitution
1.25 × V1 = 0.838 × 0.455
Rearrange to make V1 the subject
V1=

Answer:
░░░░░▐▀█▀▌░░░░▀█▄░░░
░░░░░▐█▄█▌░░░░░░▀█▄░░
░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░░
░░░░▄▄▄██▀▀▀▀░░░░░░░
░░░█▀▄▄▄█░▀▀░░
░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob
▄░▐░░░▄▄░█░▀▀ ░░
▀█▌░░░▄░▀█▀░▀ ░░ Copy And Paste Him onto all of ur brainly answers
░░░░░░░▄▄▐▌▄▄░░░ So, He Can Take
░░░░░░░▀███▀█░▄░░ Over brainly
░░░░░░▐▌▀▄▀▄▀▐▄░░
░░░░░░▐▀░░░░░░▐▌░░
░░░░░░█░░░░░░░░█
Explanation:
Answer:
Q = 30284.88 j
Explanation:
Given data:
Mass of ethanol = 257 g
Cp = 2.4 j/g.°C
Chnage in temperature = ΔT = 49.1°C
Heat required = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = 257 g× 2.4 j/g.°C × 49.1 °C
Q = 30284.88 j
"Electric charge: the proton and electron are electrically charged, while the neutron is not. The proton and electron, however, are oppositely charged. Role in the atom: Protons and neutrons are closely bound together in the nucleus of an atom, while electrons are spread out around the nucleus."