Answer:
n₁ = 1.0× 10⁻⁴ mol
Explanation:
Given data:
Initial volume of balloon = 230 mL
Initial number of moles of He =?
Final number of moles of He = 3.8 × 10⁻⁴ mol
Final volume of balloon = 860 mL
Solution:
The given problem will be solve through Avogadro law,
"Number of moles of gas and volume are directly proportional to each other at constant temperature and constant pressure"
Mathematical relationship:
V₁/n₁ = V₂/n₂
No we will put the values.
230 mL /n₁ = 860 mL/ 3.8 × 10⁻⁴ mol
n₁ = 230 mL× 3.8 × 10⁻⁴ mol/ 860 mL
n₁ = 874 × 10⁻⁴ mol. mL / 860 mL
n₁ = 1.0× 10⁻⁴ mol
Answer:
(CH2)3CH3 > CH2CH2CH3 > CH2CH3 > CH3
Explanation:
Giving the following ; CH3, CH2CH3, CH2CH2CH3, (CH2)3CH3
Priority increases as the number of CH2 group increases and vice versa, as such the one with more CH2 group will be the highest priority and the least compound with the small CH2 group attached, will have the smallest priority.
The arrangement is as follows ; (CH2)3CH3 > CH2CH2CH3 > CH2CH3 > CH3
Answer:
The answer to your question is: letter E, 1.77 L of 1.55 M Ba(NO3)2
Explanation:
Formula
Molarity = 
moles = Molarity x volume (L)
a. 2.58 L of 0.0250 M Ba(NO3)2
moles = (0.025) (2.58)
moles = 0.065 M
b. 19.23 mL of 8.5 × 10−2 M Ba(NO3)2
moles = (8.5 x 10 ⁻²) (0.01923)
moles = 0.0016
c. No right answer.
d. 26.20 mL of 2.21 M Ba(NO3)2
moles = (2.21)(0.0262)
moles = 0.058
e. 1.77 L of 1.55 M Ba(NO3)2
moles = (1.55)(1.77)
moles = 2.74 This solution needs the greatest concentration of
Ba(NO₃)₂
The nucleus, that dense central core of the atom, contains both protons and neutrons. Electrons are outside the nucleus in energy levels. Protons have a positive charge,neutrons have no charge, and electrons have a negative charge. A neutral atom contains equal numbers of protons and electrons.
Answer:
∴ Q = -7.52kCal
Explanation:
Using the formula for specific heat capacity:
Q = mcΔT
where ΔT = change in temperature (final - initial) = (0 - 100)°C = -100°C
m = mass (g) = 75g
c = specific heat capacity = 4.2 J/g°C in water
⇒ Q = 75 × 4.2 × -100
= -31,500J
But 1J - 0.000239kCal
<u>∴ Q = -7.52kCal</u>
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Let me know if I can be of further assistance.
