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vitfil [10]
3 years ago
11

150 grams of borax needs to be dissolved in water. Even after the solution is stirred, some borax remains visible in the water.

What can we add to this solution to help dissolve the remaining borax quicker?
Chemistry
1 answer:
tatyana61 [14]3 years ago
3 0
More solvent (water)
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Please help! i will give brainliest!
Molodets [167]

Answer:

i think this would be an example of a physical change because here the substance is changing directly from a solid to a gas without passing through the liquid state which means that the composition of the substance has not changed and remains the same. This is also called sublimation.

6 0
2 years ago
A student wishes to prepare 25mL of 0.1M NaOH from 6M NaOH.
liraira [26]

Answer:

Na(OH)4

Explanation:

Look at the charges and add them up

Na(OH) Na(OH)

7 0
3 years ago
Write the chemical equation for the reaction of ammonia with hydrogen chloride
raketka [301]
If you mix Ammonia as well as Hydrogen chloride, you will get ammonium chloride. It is given by NH3 (g) + HCl (g) → NH4Cl(s).
3 0
3 years ago
AuB4O7 name<br> Element i need help
coldgirl [10]
Can you ask your question more thoroughly?
7 0
3 years ago
What is the energy required to change a spherical drop of water to five smaller spherical drops of equal size? At room temperatu
pav-90 [236]
T<span>his is a straightforward question related to the surface energy of the droplet. </span>

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
3 0
3 years ago
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