Answer:
The true statement is option A.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 1 atm
V = Volume of gas = ?
n = number of moles of gas = 1 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 273.15 K
V = 22.42 L
This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.
So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.
0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure
0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Hey there!:
Molar mass Ca(NO2)2 = 132.089 g/mol
Mass of solute = 120 g
Number of moles:
n = mass of solute / molar mass
n = 120 / 132.089
n = 0.0009084 moles of Ca(NO2)2
Volume in liters of solution :
240 mL / 1000 => 0.24 L
Therefore:
Molarity = number of moles / volume of solution
Molarity = 0.0009084 / 0.24
Molarity = 0.003785 M
Hope that helps!
Explanation:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
The atomic mass of isotope (II) of vanadium is 49.944 amu.
