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3 years ago

which of the following will stay constant,no matter is the substance is in a solid,liquid,or gas states

2 answers:

7 0

Gas will stay constant no matter the substance because gas is the only state where things can shrink to fit a tiny box or stretch out to fill a whole room.

Vlada [557]3 years ago

6 0

I believe it is a gas

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Structure can affect the Ka values for related acids. In the boxes below, draw the complete Lewis structure for a single ion of

Zigmanuir [339]

Answer:

Explanation: see attachment below

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3 years ago

If all the deer disappeared from this community, which change would be most

butalik [34]

Answer:c

Explanation:

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2 years ago

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When the products of a reaction have less energy than the reactants

ohaa [14]

Answer:

the energy of the products is less than the energy of the reactants.

Explanation:

the the change is enthalpy is negative, and heat is released to the surroundings.

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2 years ago

calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.

Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

$P_w=P_p+P_s$

$P_w=P_p+(-iCRT)$

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= $P_w$

$P_p=0 bar$ (At standard temperature)

$P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar$

$P_w=0 bar+(-3.406 ) bar$

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

7 0

2 years ago

One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

5 0

2 years ago