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luda_lava [24]
3 years ago
13

The data part of Lab Hydrates in Edgunuity

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
3 0

Answer:

Hydrates! :D

Explanation:

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If the initial volume is 55ml and the final volume after some rocks of sandstone are added is 65ml what is the volume of the chi
AfilCa [17]

Answer:

Volume = 10ml

Density = 1/5 g/ml or 0.20g/ml

Explanation:

The rocks are 10ml since the initial volume went up by 10.

Since density = mass/volume, you divide 2 by 10.

D = 2/10

D = 1/5 g/ml or 0.20g/ml

(Unit is g/ml aka grams/millileter)

7 0
3 years ago
if the density of a bar of gold is 19.3g/cm^3and you cut it into four pieces , what is its density of each piece of gold
AlladinOne [14]
Density is an intrinsic property, so it is independent of the amount of substance present: one gold coin would have the same density as a solid gold boulder.

So if the density of gold is 19.3 g/cm³, the density of a bar of gold and the pieces into which the bar is cut would all be 19.3 g/cm³.
4 0
4 years ago
Giving brainly if correct​
Kamila [148]

Answer:

D

Explanation:

filtration, the process in which solid particles in a liquid or gaseous fluid are removed by the use of a filter medium that permits the fluid to pass through but retains the solid particles.

Hope that helps

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6 0
2 years ago
Read 2 more answers
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
3 years ago
2.41 x 10^2 Cm to meters
Assoli18 [71]

We know that there are 100 cm in 1 m, so we can use this to convert to meters:

(\frac{2.41x10^{2}cm}{1})*(\frac{1m}{100cm} )=2.41m

Therefore we know that 2.41x10^{2} cm is equal to 2.41 m.

8 0
3 years ago
Read 2 more answers
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