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2 years ago

Which has greater internal energy, 1 kg of water or 1 g of water, at the same temperature?

Chemistry

1 answer:

Natali5045456 [20]2 years ago

4 0

1kg of water has greater internal energy compared to 1g of water because 1kg of water has more mass.

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Science fair ideas? please help

egoroff_w [7]

You can Maybe do a volcano?

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3 years ago

Read 2 more answers

In the equation:

Ksenya-84 [330]

Answer:

B) 16 g

Explanation:

2H₂ + O₂ → 2H₂O

First we convert 4 moles of O₂ into moles of H₂, using the stoichiometric coefficients of the balanced reaction:

4 mol O₂ * $\frac{2molH_2}{1molO_2}$ = 8 mol H₂

Finally we convert 8 moles of H₂ into grams, using its molar mass:

8 mol H₂ * 2 g/mol = 16 g

Thus, the correct answer is option B).

6 0

2 years ago

An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm

frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

xA= 0.34, yA= 0.55

ii) 0.50 = 0.55 nv + 0.34 nL

Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0

2 years ago

Why can filtration not be used to seperate a mixture of 2 liquids

Salsk061 [2.6K]

Answer:

Filtration works best when the solute isn't dissolve in the solvent. For instance, sand and water can be seperated through filtration because both compounds do not dissolve with each other. However, sugar and water would not be seperated through filtration as they dissolve with each other.

-google :)

3 0

3 years ago

4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a

yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0

2 years ago