Answer:
The answer to the question is
The specific heat capacity of the alloy = 1.77 J/(g·°C)
Explanation:
To solve this, we list out the given variables thus
Mass of alloy = 45 g
Initial temperature of the alloy = 25 °C
Final temperature of the alloy = 37 °C
Heat absorbed by the alloy = 956 J
Thus we have
ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g
∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or
c = 956 J/(540 g·°C) = 1.77 J/(g·°C)
The specific heat capacity of the alloy is 1.77 J/(g·°C)
Answer: 300 K
Explanation:
Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.
(At constant pressure and number of moles)

Given : V= 6.0 L
k= 0.020 L/K
T=?


Thus temperature of the gas is 300 K.
Answer:
3.72 mol Hg
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Density = Mass over Volume
Explanation:
<u>Step 1: Define</u>
D = 13.6 g/mL
54.8 mL Hg
<u>Step 2: Identify Conversions</u>
Molar Mass of Hg - 200.59 g/mol
<u>Step 3: Find</u>
13.6 g/mL = x g / 54.8 mL
x = 745.28 g Hg
<u>Step 4: Convert</u>
<u />
= 3.71544 mol Hg
<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.71544 mol Hg ≈ 3.72 mol Hg
Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient: ![Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2](https://tex.z-dn.net/?f=Q_%7Br%7D%20%3D%5Cfrac%7B%5Cleft%20%5B%20PEP%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%202PG%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B0.1%20mM%7D%7B0.5%20mM%7D%20%3D%200.2)
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

![\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20kJ%2Fmol%20%2B%20%5B2.303%20%5Ctimes%20%288.314%20%5Ctimes%2010%5E%7B-3%7D%20kJ%2F%28K.mol%29%29%5Ctimes%20%28310.15%20K%29%5D%20log%20%280.2%29)
![\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20%2B%20%5B5.938%5D%20%5Ctimes%20%28-0.699%29%20%3D%201.7%20-%204.15%20%3D%20%28-2.45%20kJ%2Fmol%29)
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>
The answer is heterogeneous mixture<span> because the </span>blood<span> cells are physically separate from the </span>blood<span> plasma.</span>