Answer:
Check all of them
Explanation:
All are correct.
- stong electrolytes dissolve completely
- weak disolve partially
- nonelectrolytes do not dissolve
Answer:
Human ears can hear sound waves that vibrate in the range from about 20 times a second (a deep rumbling noise) to about 20,000 times a second (a high-pitched whistling). (Children can generally hear higher-pitched sounds than their parents, because our ability to hear high frequencies gets worse as we get older.) Speaking more scientifically, we could say that the sounds we can perceive have a frequency ranging from 20–20,000 hertz (Hz). A hertz is a measurement of how often something vibrates and 1 Hz is equal to one vibration each second. The human voice makes sounds ranging from a few hundred hertz to a few thousand hertz.
Suppose you could somehow hit a drum-skin so often that it vibrated more than 20,000 times per second. You might be able to see the skin vibrating (just), but you certainly couldn't hear it. No matter how hard you hit the drum, you wouldn't hear a sound. The drum would still be transmitting sound waves, but your ears wouldn't be able to recognize them. Bats, dogs, dolphins, and moths might well hear them, however. Sounds this like, with frequencies beyond the range of human hearing, are examples of ultrasound.
Infrasonics, vibrational or stress waves in elastic media, having a frequency below those of sound waves that can be detected by the human ear—i.e., below 20 hertz. The range of frequencies extends down to geologic vibrations that complete one cycle in 100 seconds or longer.
Answer:
Theanswer to your question is:
Limiting reactant = FeCl₃
Excess reactant = 1.66 g of Mg
Explanation:
Data
Mg = 41 g = 24.31 g/mol
FeCl₃ = 175 g = 162.2 g/mol
3Mg(s) + 2FeCl₃(s) → 3MgCl₂(s) + 2Fe(s)
3(24.31) of Mg ------------------ 2(162.2) of FeCl₃
72.93 g of Mg ------------------ 324.4 g of FeCl₃
Theoretical Proportion = 324.4/72.93 = 4.44
Practical proportion = 175 / 41 = 4.2
As the proportion disminishes the limiting reactant is FeCl₃.
Excess reactant
72.93 g of Mg ------------------ 324.4 g of FeCl₃
x ------------------------- 175 g of FeCl₃
x = (175 x 72.93) / 324.4
x = 39.34 g of Mg
Excess = 41 - 39.34
= 1.66 g of Mg
Answer:
answer it by yourself and own way
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
