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My name is Ann [436]
3 years ago
14

60 grams of ice will require _____ calories to raise the temperature 1°C.

Chemistry
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

<em>It is known that 1.0 cal = 4.18 J.</em>

<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>

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Fill in the chart to describe and give examples of physical changes.
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Answer:

here are some examples of physical change!!!

Explanation:

-An ice cube melting into water in your drink.

-Freezing water to make ice cubes.

-Boiling water evaporating.

-Hot shower water turning to steam.

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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment
astraxan [27]

Answer:

The activation energy is =8.1\,kcal\,mol^{-1}

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_{A}=kC_{A}

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^{3}

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_{o}=5dm^{3}s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]

Rearrange the formula is as follows.

k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)

The feed has Pure A, mole fraction of A in feed y_{A_{o}} is 1.

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

Substitute the all given values in equation (1)

k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}

Therefore, the rate constant in case of the plug flow reacor at 300K is1.2s^{-1}

The rate constant in case of the CSTR can be calculated by using the formula.

\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed y_{A_{o}} is 0.5

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacted.

=0.5(2-1)=0.5

Substitute the all values in formula (2)

\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}

Therefore, the rate constant in case of CSTR comes out to be 2.8s^{-1}

The activation energy of the reaction can be calculated by using formula

k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}

Substitute the all values.

=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}

=8.1\,kcal\,mol^{-1}

Therefore, the activation energy is =8.1\,kcal\,mol^{-1}

8 0
3 years ago
3. What is the mass of HI used to create 2.50L of a 0.48 M solution of hydroiodic acid?
Daniel [21]

Answer:

Mass = 153.48 g

Explanation:

Given data:

Volume of solution = 2.50 L

Molarity = 0.48 M

Mass required = ?

Solution:

Molarity = number of moles / volume in litter

Number of moles = Molarity × volume in litter

Number of moles = 0.48 M  × 2.50 L

Number of moles =  1.2 mol

Mass of HI:

Number of moles = mass/molar mass

Mass = Number of moles × molar mass

Mass = 1.2 mol × 127.9 g/mol

Mass = 153.48 g

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