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mariarad [96]
3 years ago
5

What is the mass of a sample of a sample of iron (cp=0.44j/g•°C) when it is heated from 60°to 160°C. The heat is transferred is

6600j.
Chemistry
1 answer:
scZoUnD [109]3 years ago
5 0

Q=mcat

6600=(m)(.44)(100)

6600=(44m)

150= m

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7 0
3 years ago
How does a catalyst increase the rate of a reaction?
Norma-Jean [14]

Answer:

C) It provides a lower activation energy for the reaction is the correct answer.

Explanation:

  • A catalyst increases the rate of the chemical reaction by lowering the activation energy for a reaction.
  • Catalyst is used to increase the reaction rate, it remains unchanged in the chemical reaction and it does not change the equilibrium constant.
  • Activation energy is a minimum amount of energy required to initiate the reaction.

6 0
3 years ago
How many carbon atoms are there in a diamond (pure carbon) with a mass of 55 mg ?
kari74 [83]
1 mol of Carbon = 12 grams.
x mol of Carbon = 55 grams

12*x = 1 * 55
x = 55/12
x = 4.583333 mols of carbon

1 mol of anything is 6.02 * 10^23 atoms
4.58333333 mol = x

1/4.5833333 = 6.02 * 10^23/x
x = 4.58333* 6.02*10^23
x = 2.7591 * 10^23 Carbon atoms
3 0
3 years ago
If the heating curve is reversed, what describes the boiling point?
GalinKa [24]
Boiling is the process of converting a substance from liquid state to gaseous state. If the heating curve is reversed, the process also is reversed from converting gaseous state to liquid state. In this case, the reverse of boiling is condensation. So the answer is point of condensation.
5 0
3 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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