Says here the answer is <span>dipole-dipole</span>
Explanation:
Given that,
Mass of metal cube, m = 20 g
Volume of cube, V =5 mL
We need to find the density of the cube. Mass per unit volume equals density.
![d=\dfrac{m}{V}\\\\d=\dfrac{20\ g}{5\ mL}\\\\d=4\ g/mL](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7Bm%7D%7BV%7D%5C%5C%5C%5Cd%3D%5Cdfrac%7B20%5C%20g%7D%7B5%5C%20mL%7D%5C%5C%5C%5Cd%3D4%5C%20g%2FmL)
So, the density of the cube is 4 g/ml.
We know that the density of Aluminium is 4 g/mL.
So, the cube is not Aluminium cube.
Answer:
The correct answer is - 800.
Explanation:
Given:
Total amount = ? or assume x
spend in buying birthday item = 3/4 of x
given to sister = 1/5 of x
remaining to mother = 40
solution:
the remaning amount = x- (3x/4+x/5) = 4=
=> x- 19x/20 = 40
=> x = 20*40
=> x = 800
thus, the correct answer is = 800
The correct answer attached in file, Thank you for joining brainly community.
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.