If the change in Gibbs free energy for a process is positive, the corresponding change in entropy for the universe will be positive.
<h3>What is Gibbs free energy?</h3>
This is defined as the energy used by a substance involved in a chemical reaction.
The Gibbs free energy and entropy has a direct relationship which is why a positive gibbs free energy will result in a corresponding positive entropy.
Read more about Gibbs free energy here brainly.com/question/9179942
Answer:
it states that the total mass of the products are the same as the total mass of the reactants in a chemical reaction.
Explanation:
Answer:
0.19 g
Explanation:
Step 1: Given data
Volume of hydrogen at standard temperature and pressure (STP): 2.1 L
Step 2: Calculate the moles corresponding to 2.1 L of hydrogen at STP
At STP (273.15 K and 1 atm), 1 mole of hydrogen has a volume of 22.4 L if we treat it as an ideal gas.
2.1 L × 1 mol/22.4 L = 0.094 mol
Step 3: Calculate the mass corresponding to 0.094 moles of hydrogen
The molar mass of hydrogen is 2.02 g/mol.
0.094 mol × 2.02 g/mol = 0.19 g
Answer:
1.2* 10³ rNe.
Explanation:
Given speed of neon=350 m/s
Un-certainity in speed= (0.01/100) *350 =0.035 m/s
As per heisenberg uncertainity principle
Δx*mΔv ≥\frac{h}{4\pi }
4π
h
..................(1)
mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg
6.22∗10
−23
20∗10
−3
=3.35∗10
−26
kg
substituating the values in eq. (1)
Δx =4.49*10^{-8}10
−8
m
In terms of rNe i.e 38 pm= 38*10^{-12}10
−12
Δx=\frac{4.49*10^{-8} }{38*10^{-12} }
38∗10
−12
4.49∗10
−8
=0.118*10^{4}10
4
* (rNe)
=1.18*10³ rN
= 1.2* 10³ rNe.
Explanation:
This is the answer
Answer:
<span>ρ≅13.0⋅g⋅m<span>L<span>−1</span></span></span> = <span>13.0⋅g⋅c<span>m<span>−3</span></span></span>
Explanation:
<span>Density=<span>MassPer unit Volume</span></span> = <span><span>75.0⋅g</span><span><span>(36.5−31.4)</span>⋅mL</span></span> <span>=??g⋅m<span>L<span>−1</span></span></span>
Note that <span>1⋅mL</span> = <span>1⋅c<span>m<span>−3</span></span></span>; these are equivalent units of volume;
i.e. <span>1⋅c<span>m3</span></span> = <span>1×<span><span>(<span>10<span>−2</span></span>⋅m)</span>3</span>=1×<span>10<span>−6</span></span>⋅<span>m3</span>=<span>10<span>−3</span></span>⋅L=1⋅mL</span>.
