Answer:
The answer is
<h2>2.71 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
volume of marble = 1564 cm³
1 kg = 1000 g
4.24 kg = 4240 g
mass = 4240 g
The density is
We have the final answer as
<h3>2.71 g/cm³</h3>
Hope this helps you
Answer:u would have 40
Explanation:
because ur taking them away
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
Answer:
The heat of vaporization is typically larger than the heat of fusion
Next question answer:
The liquid water absorbs heat from the skin surface and is transferred to the air when the water evaporates.
Explanation:
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
