Answer:
64,433.6 Joules
Explanation:
<u>We are given</u>;
- Volume of water as 220 mL
- Initial temperature as 30°C
- Final temperature as 100°C
- Specific heat capacity of water as 4.184 J/g°C
We are required to calculate the amount of heat required to raise the temperature.
- We know that amount of heat is calculated by;
Q = mcΔT , where m is the mass, c is the specific heat, ΔT is the change in temperature.
Density of water is 1 g/mL
Thus, mass of water is 220 g
ΔT = 100°C - 30°C
= 70°C
Therefore;
Amount of heat, Q = 220g × 4.184 J/g°C × 70°C
= 64,433.6 Joules
Thus, the amount of heat required to raise the temperature of water is 64,433.6 Joules
Answer:
4.375 milligram
Explanation:
Applying,
A = A'(2ᵃ/ⁿ)............. Equation 1
Where A = original amount of actinium-228, A' = amount of actinium-228 left after decay, a = Total time, n = half life
make A' the subject of the equation
A' = A/(2ᵃ/ⁿ)........... Equation 2
From the question,
Given: A = 70 milligram, a = 24.52 hours, n = 6.13 hours
Substitute these values into equation 2
A' = 70/(2²⁴°⁵²/⁶°¹³)
A' = 70/(2⁴)
A' = 70/16
A' = 4.375 milligram
The best way to accurately determine the pair with the highest electronegativity difference is by using their corresponding electronegativity values. For the each of the choices, the difference is:
A. H-S = 2.5 - 2.1 = 0.4
B. H-Cl = 3 - 2.1 = 0.9
C. N-H = 3 - 2.1 = 0.9
D. O-H = 3.5 - 2.1 = 1.4
E. C-H = 2.5 - 2.1 = 0.4
As show, D. has the highest difference. Without looking at their values, you can also determine the pair with the highest difference by taking note of the trend of electronegativity on the periodic table. Electronegativity increases as you go right a group and up a period. This makes oxygen the most electronegative element among the other elements paired with hydrogen.
Answer:
The answer to your question is 23.4 moles of Glucose
Explanation:
Data
moles of Oxygen = ?
moles of glucose = 3.9
-Balanced chemical reaction
1C₆H₁₂O₆ + 6O₂ ⇒ 6H₂O + 6CO₂
Process
1.- To solve this problem, use the coefficient of the balanced chemical equation, and use proportions and cross multiplication.
1 mol of C₆H₁₂O₆ -------------------- 6 moles of O₂
3.9 moles of C₆H₁₂O₆ ------------------ x
x = (3.9 x 6 ) / 1
x = 23.4 moles
2.- Conclusion
3.9 moles of glucose consume 23.4 moles of oxygen
