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3 years ago

Which term best describes liquid behavior under pressure?

Chemistry

1 answer:

Lelechka [254]3 years ago

6 0

Answer:

The correct answer is incompressible. Liquids or fluids that behave under pressure are called incompressible fluids because they tend to behave through flowing movements

Explanation:

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Answer:

The energy released from food was once energy from the sun that was captured by plants in the chemical process that forms plant matter (from air and water). Food provides animals with the materials they need for body repair and growth and the energy they need to maintain body warmth and for motion.

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4 years ago

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What types of materials are usually ductile

White raven [17]

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4 years ago

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How many grams of sodium hyrdioxde are in 251.0ml f 0.600m

Furkat [3]

Answer:

There are 6.024 grams of sodium hydroxide in the solution.

Explanation:

$Molarity=\frac{Moles}{Volume (L)}$

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of sodium hydroxide = n

Volume of sodium hydroxide solution = 251.0 mL = 0.251 L

Molarity of the sodium hydroxide = 0.600 M

$n=0.600 M\times 0.251 L=0.1506 mol$

Mass of 0.1506 moles of NaOH :

$40 g/mol\times 0.1506 mol = 6.024 g$

There are 6.024 grams of sodium hydroxide in the solution.

4 0

3 years ago

A mixture of so2(g) and o2(g), formed by the complete decomposition of so3(g), is collected over water at 34°c at a total pressu

Sedbober [7]

The decomposition of so3 to so2 gas and o2 gas can be described in the balanced chemical equation:

2so3(g) = 2 so2 (g) + 02(g).

so assuming a complete reaction, the ratio of so2 gas to total products is 2/3 while that of 02 is 1/3.

Subtracting water's water vapor pressure, 760-40 mm hG = 720 mm Hg.

then the products partial pressures are

so2 = 2/3 * (720) = 480 mm Hg.
o2 = 720-480 = 240 mm Hg.

8 0

3 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0

3 years ago