Explanation:
The magnet uses electromagnetic induction meaning it can be readily magnetized and demagnetized (using electricity) when required. When electricity is switched on, it becomes magnetized and lifts an object and when electricity is switched off, it loses magnetism and releases the object.
The magnetic is able to lift heavy objects because it has powerful conductor material (ferromagnetic iron) and the number of electromagnetic coils is many to induce a powerful magnetic force. The electric current, inducing the magnetism, is also powerful.
Learn More:
For more on electromagnetic induction check out;
brainly.com/question/3414535
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Answer: An electron having a quantum number of one is closer to the nucleus
Explanation:
The Bohr model relies on electrostatic attraction between the nucleus and orbital electron. Hence, the closer an electron is to the nucleus the more closely it is held by the nucleus and the lesser its energy (the more stable the electron is and the more difficult it is to ionize it). The farther an electron is from the nucleus ( in higher shells or energy levels), the less the electrostatic attraction of such electron to the nucleus due to shielding effect. Hence it is less tightly held.
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
Answer:Exothermic and endothermic reactions (ESBQM)
Explanation:
Answer:
49.5J/°C
Explanation:
The hot water lost some energy that is gained for cold water and the calorimeter.
The equation is:
Q(Hot water) = Q(Cold water) + Q(Calorimeter)
<em>Where:</em>
Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J
Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J
That means the heat gained by the calorimeter is
Q(Calorimeter) = 4794J - 3834J = 960J
The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:
59.4°C-40°C = 19.4°C
And calorimeter constant is:
960J/19.4°C =
<h3>49.5J/°C</h3>
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