52.8 grams of fat would be contained in 0.97 lbs of ground beef
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
Answer:
0.15066667
I tried, hope this helps :)
Answer:
S = 1.1 × 10⁻⁹ M
Explanation:
NaCl is a strong electrolyte that dissociates according to the following expression.
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.
We can find the molar solubility (S) of AgCl using an ICE chart.
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
I 0 0.15
C +S +S
E S 0.15+S
The solubility product (Ksp) is:
Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)
If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M
<u><em>Answer:</em></u>
<u><em>Explanation:</em></u>
Phosphate ion has covalent bond. As we known , covalent bond is formed by sharing of electrons between two atoms. As in this case, P form three single covalent bond with three oxygen atoms and one double covalent bond with one oxygen . The formal charges is -3.
While He is present in mono atomic form
NaI have ionic bond which is formed by donating electrons from one atom and other accept.
Ag is present independently with no other atom.
