How does the specific heat of water compare to the specific heat of concrete?

A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen

Elenna [48]

Using the significant figure it would be 27.3

8 0

2 years ago

An analysis of a sugar compound is found to contain 132 grams of carbon (C), 22 grams of hydrogen (H), and 176 grams of oxygen (

ddd [48]

132 g of C , 22 g of H , 176 g of O

132 + 22 + 176 => 330 g <span>of the substance

</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u

C = 132 / 12.0 => 11 moles

H = 22 / 1.0 => 22 moles

O = 176 / 16.0 => 11 moles

Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1

H = 22 / 11 => 2

O = 11 / 11 => 1

formula empirically <span>is : CH</span>₂O

hope this helps!

8 0

3 years ago

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the

Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}$

$P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}$

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

#SPJ4

4 0

1 year ago

Is there a qualitative difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process

Vesna [10]

Answer:

No

Explanation:

given that, enthalpy is a state function, that means it depends only on the initial and final states, there is no difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process, when the cooling or heating process finish in a change of phase.

It does not matter which way we take to cool or heat the substances the Enthalpy of this process will be the same.

3 0

3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

taurus [48]

Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

Atomic mass of the elements are:

Na: 22.989 u

S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

= (17.240 g)/(158.105 g/mol) = 0.1090 mole.

Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

6 0

3 years ago