We are asked to convert 25 cg to units of hg.
1 cg = 1 centigram = 10⁻² g
1 hg = 1 hectogram = 10² g
The options given are:
a) 1 hg/ 10² g
b) 10² cg/ 1 hg
c) 10² hg/ 1 cg
d) 10⁻² g/ 1 cg
To convert 25 cg to 1 hg, we could convert the 25 cg to grams first, then grams to hg.
25 cg · 10⁻² g/ 1cg = 0.25 g
Here we have converted our number from cg to grams. We can use another conversion of grams to hg to complete the conversion.
0.25 g · 1 hg/ 10² g = 0.0025 hg
Therefore, the first conversion we used was d) 10⁻² g/ 1 cg.
<h2>
Hello!</h2>
The answer is:
The new volume will be 1 L.
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Why?</h2>
To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.
To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.
Boyle's Law equation is:
So, we are given the information:
Then, isolating the new volume and substituting into the equation, we have:
Hence, the new volume will be 1 L.
Have a nice day!
The alpha particles that were fired at the gold foil were positively charged. ... These experiments led Rutherford to describe the atom as containing mostly empty space, with a very small, dense, positively charged nucleus at the center, which contained most of the mass of the atom, with the electrons orbiting the nucleus.
hope this helped
Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J
ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
