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Snezhnost [94]
3 years ago
10

A thermometer that was used in a calorimeter was consistently 2.7 °c too high. how would this affect the calculations of δh for

the reactions conducted in the calorimeter?
Chemistry
1 answer:
vodomira [7]3 years ago
7 0
The calorimetry experiment is usually done to determine the heat capacity of the sample. The working equation would be:

Q = mCpΔT
Q is the energy
m is the mass of sample
Cp is the heat capacity
ΔT is the temperature

So, if the thermometer is too high, then that would affect ΔT, which would make it greater. Consequently, you would calculate a much lesser heat capacity of the sample compared to the theoretical value.
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Which law relates to the ideal gas law?
Kamila [148]

The equation of state for a hypothetical ideal gas is known as the ideal gas law, sometimes known as the general gas equation. i.e. PV = nRT or P1V1 = P2V2.

  • According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.
  • Robert Boyle, Gay-Lussac, and Amedeo Avogadro's observational work served as the basis for the ideal gas law. The Ideal gas equation, which simultaneously describes every relationship, is obtained by combining all of their observations into a single statement.
  • When applying the gas constant R = 0.082 L.atm/K.mol, pressure, volume, and temperature should all be expressed in units of atmospheres (atm), litres (L), and kelvin (K).
  • At high pressure and low temperature, the ideal gas law basically fails because molecule size and intermolecular forces are no longer negligible but rather become significant considerations.

Learn more about ideal gas law here:

brainly.com/question/26040104

#SPJ9

3 0
1 year ago
Calculate the number of milliliters of 0.710 M Ba(OH)2 required to precipitate all of the Mn2+ ions in 161 mL of 0.796 M KMnO4 s
USPshnik [31]

<u>Answer:</u> The volume of barium hydroxide is 183 mL.

<u>Explanation:</u>

To calculate the moles of cadmium nitrate, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}       .....(1)

Molarity of MnSO_4 = 0.796 M

Volume of MnSO_4 = 161 mL = 0.161 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)

By Stoichiometry of the reaction:

1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.

So, 0.13 moles of manganese sulfate will react with = \frac{1}{1}\times 0.13=0.13mol of barium hydroxide

Now, calculating the volume of barium hydroxide by using equation 1, we get:

Moles of barium hydroxide = 0.13 moles

Molarity of barium hydroxide = 0.710 M

Putting values in equation 1, we get:

0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So, 0.183L=0.183\times 1000=183mL

Hence, the volume of barium hydroxide is 183 mL.

7 0
3 years ago
Hospital patients are administered oxygen from an pressurized
VashaNatasha [74]

Answer:

32000atm

Explanation:

Using Boyle's law equation;

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (

V2 = final volume (L)

According to the question below:

P1 = 160.0 atm

P2 = 3.0 atm

V1 = 600L

V2 = ?

Using P1V1 = P2V2

160 × 600 = 3 × V2

96000 = 3V2

V2 = 96000/3

V2 = 32000atm

5 0
3 years ago
PLEASE HELP!!!!!
Free_Kalibri [48]
Umm I think it might be a b c in order?
4 0
2 years ago
Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25
Vladimir79 [104]

Answer:

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

  • We cam calculate the solubility of O₂ in water using Henry's law: <em>Cgas = K P</em>,
  • where, Cgas is the solubility if gas,
  • K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),
  • P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).
  • Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

2) The effect of decreasing temperature on the solubility O₂ gas in water:

  • Decreasing the temperature will increase the solubilty of O₂ gas in water.
  • When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.
  • Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.


6 0
3 years ago
Read 2 more answers
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