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zmey [24]
2 years ago
14

How much heat has been lost when a 78 g piece of copper is placed into a calorimeter and cools from 120°C to

Chemistry
1 answer:
lora16 [44]2 years ago
3 0

The amount of heat lost by the copper is 2402.4 J.

To Calculate the amount of heat lost, we use the formula below.

<h3>Formula:</h3>
  • Q = cm(t₂-t₁)................. Equation 1

<h3>Where:</h3>
  • Q = Amount of heat lost
  • c = specific heat capacity of copper
  • m = mass of copper
  • t₂ = Final temperature
  • t₁ = Initial temperature

From the question,

<h3>Given:</h3>
  • m = 78 g
  • c = 0.385 J/g°C
  • t₂ = 120°C
  • t₁ = 40°C

Substitute these values into equation 1.

  • Q = 78(0.385)(120-40)
  • Q = 2402.4 J

Hence, The heat lost by the copper is 2402.4 J

Learn more about heat here: brainly.com/question/13439286

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Substance P replaces X in the compound XY

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6 0
3 years ago
The mass composition of a compound that assists in the coagulation of blood is 76.71% carbon, 7.02% hydrogen, and 16.27% nitroge
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Answer : The empirical of the compound is, C_{11}H_{12}N_2

Explanation :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 76.71 g

Mass of H = 7.02 g

Mass of N = 16.27 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{76.71g}{12g/mole}=6.39moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.02g}{1g/mole}=7.02moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.27g}{14g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{6.39}{1.16}=5.5

For H = \frac{7.02}{1.16}=6.0\approx 6

For N = \frac{1.16}{1.16}=1

The ratio of C : H : N = 5.5 : 6 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : N = 11 : 12 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_{11}H_{12}N_2

Therefore, the empirical of the compound is, C_{11}H_{12}N_2

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A family ppoo holds 10000 gallons of water. how many cubic meters is this
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The enzyme Y catalyzes the elementary reaction
lions [1.4K]

Answer:

0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM

Explanation:

An enzyme solution is saturated when all the active sites of the enzyme molecule are full.  When an enzyme solution is saturated, the reaction is occurring at the maximum rate.

From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.

The addition of more substrate to a solution that contains the enzyme required  for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.

<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of  0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>

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