B. Exosphere
Prefix "exo" means outside, or outer. So, the exosphere is the outer layer of the Earth, which is not a part of the water cycle.
There are three significant figures, this is because the zero before the decimal does not count
Answer:
The mass of copper sulfate in 25 cm³ of the copper sulfate solution is 0.45 g
Explanation:
The given parameters are;
The volume of the copper sulfate solution = 100 cm³
The mass of the copper sulfate in the solution = 1.8 g
Therefore, the mass of copper sulfate in 25 cm³ of the solution is given as follows;
The mass of copper sulfate in 100 cm³ of the solution = 1.8 g
The mass of copper sulfate in 1 cm³ of the solution = 1.8 g/100 = 0.018 g
Therefore;
The mass of copper sulfate in 25 × 1 = 25 cm³ of the solution, m = 25×0.018 g = 0.45 g
∴ The mass of copper sulfate in 25 cm³ of the solution, m = 0.45 g
<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
This is an incomplete question, here is a complete question.
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem. If the change in enthalpy is -5074.2 kJ, how much work is done during the combustion? Express the work in kilojoules to three significant figures.
Answer : The work done during the combustion is, 9.9 kJ
Explanation :
Formula used :
where,
w = work done = ?
= change in enthalpy = -5074.2 kJ
= change in internal energy = -5084.1 kJ
R = gas constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
Thus, the work done during the combustion is, 9.9 kJ