<span>According to my knowledge, I feel the answer is -
Particles that struck the center of the atom were repelled.
Hope this helps!
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Answer: 64.6 mmHg
Explanation:
Given that:
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
Recall that pressure of the gas is required in mm hg, so convert 0.085 atm to mm Hg
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
Thus, the pressure of the gas is 64.6 mm hg
Yes the answer is 1s22s22p63s23p64s23d5
Where is the table? I dont know what to classify
Answer:
660kcal
Explanation:
The question is missing the concentration of the glucose solution. Standard glucose concentration for IV solution is 5% or 5g of glucose every 100mL of solution.
We need to determine how many grams of glucose are there inside the solution. The number of glucose in 3.3L solution will be:
3.3L * (1000mL / L) * (5g/100mL)= 165 g.
If glucose will give 4kcal/ g, then the total calories 165g glucose give will be: 165g * 4kcal/ g= 660kcal.
