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mote1985 [20]
3 years ago
7

(Math) I NEED THIS ANSWER PLZ

Chemistry
1 answer:
WARRIOR [948]3 years ago
6 0

Answer:

Q

Explanation:

√10=3.16

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What are the signs of the enthalpy change (ΔH°) and the entropy change (ΔS°) for the condensation of CS2(g)?
VARVARA [1.3K]

Answer:

∆H is negative

∆S is negative

Explanation:

The condensation of CS2 implies a phase change from gaseous state to liquid state. The energy of the gaseous particles is greater than that of the liquid particles hence energy is given out when a substance changes from gaseous state to liquid state hence the process is exothermic and ∆H is negative.

Changing from gaseous state to liquid states leads to a decrease in entropy hence ∆S is negative. Liquid particles are more orderly than particles of a gas.

5 0
4 years ago
WILL GIVE BRAINLY
worty [1.4K]

In order to compute mean, you have to add all values of the data set up and divide by the number of data values there are.

In this case, the total sum of all data values is: 81+88+69+65+87=390

The number of data values is 5

To compute mean: 390/5=78

The mean height of the corn stalks is 78cm (C)

3 0
3 years ago
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Which change absorbs heat?
tekilochka [14]
Freezing water into ice
7 0
3 years ago
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At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.
natka813 [3]

Explanation:

The given data is as follows.

             T = 298 K,          \Delta H^{o} = -5645 kJ/mol

          \Delta G^{o} = -5798 kJ/mol

Relation between \Delta H and \Delta G are as follows.

          \Delta G^{o} = \Delta H^{o} - T \Delta S^{o}    

             -5798 kJ/mol = -5645 kJ/mol - 298 \times \Delta S^{o}

                       -153 kJ/mol = -298 \times \Delta S^{o}

                    \Delta S^{o} = 0.513 kJ/mol K

Now, temperature is 37^{o}C = (37 + 273) K = 310 K

Since,        \Delta G = \Delta H^{o} - T \Delta S^{o}

                            = -5645 kJ/mol - 310 K \times 0.513 kJ/mol K

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            \Delta G = \Delta G_{310 K} - \Delta G_{298 K}

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0
3 years ago
Is the temperature of absolute zero measured in celsius A.-373 B.-73 C.-173 D.-273
Andrew [12]

Answer:

Option D is correct = -273 °C

Explanation:

Data given:

Temperature of absolute zero

Absolute Zero in Celsius = ?

Solution:

As we know

internationally the temperature of Absolute zero on kelvin scale  = 0 K

So to convert Kelvin temperature to degree Celsius formula will used

                  T(K) = T(°C) + 273

Rearrange the above equation for °C

                  T(°C) = T(K) - 273 . . . . (1)

Put value in above eq.1

                  T(°C) = 0 - 273

                  T(°C) =  -273

So,

The absolute zero in °C = -273

so option D is correct

3 0
3 years ago
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