It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.
The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.
<h3>How do we calculate atomic mass?</h3>
Atomic mass (A) of any atom will be calculated as:
A = mass of protons + mass of neutrons
In the Thorium-234:
Number of protons = 90
Number of neutrons = 144
Mass of one proton = 1.00728 amu
Mass of one neutron = 1.00866 amu
Mass of thorium-234 = 90(1.00728) + 144(1.00866)
Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu
Given mass of thorium-234 = 234.04360 amu
Mass defect = 235.90224 - 234.04360 = 1.85864 amu
Hence required value is 1.85864 amu.
To know more about Atomic mass (A), visit the below link:
brainly.com/question/801533
Answer:
The signal from the deceleration sensor ignites the gas-generator mixture by an electrical impulse, creating the high-temperature condition necessary for NaN3 to decompose. The nitrogen gas that is generated then fills the airbag.
basically, the nitrogen fills the bag
Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
