Which statement best describes the temperature of the ocean's surface water?

Complete the conceptap below . About Chemical Bonds

Irina18 [472]

Answer:

1. ionic bonds

2. metallic bonds

3. share

4. metal

5. non-metal

6. metals

7. NaCl ( sodium chloride )

8. CO2 ( carbon dioxide )

9. Cu ( copper )

i hope it helped......

5 0

2 years ago

3. The scientific study of how organisms are classified is called _____.

Vilka [71]

Hey mate!

Taxonomy is the study of classification of organisms. Therefore, your answer is A.

Hope this helps!

3 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Question 25 of 33

Dafna11 [192]

Answer:

A) It tends to increase from top to bottom of a group

6 0

3 years ago

C6H12O6(s) + 6O2(g) --> 6H2O(g) + 6CO2(g)

vredina [299]

One
Balance the Equation.
This has been done for you or it is given. Anyway this step is finished, but it must always be done.

Two
Find the molar mass of C6H12O6
6C = 6 * 12 = 72
12H = 12*1 = 12
CO = 6 * 16 = 96
Mol Mass = 180 grams / mol

Three
Find the mols of C6H12O6
n = ???
Molar Mass = 180 grams / mol
given mass = 13.2 grams.

n = given mass / molar mass
n = 13.2 / 180
n = 0.07333333 mols.

Four
Find the mols CO2
1 mol C6H12O6 will produce 6 mols CO2
0.0733333 mols will produce x

1/0.073333 = 6/x Cross multiply
x = 0.073333 * 6
x = 0.4399 moles.

Five
Find the volume given the conditions for temperature and pressure are STP conditions.
PV = nRT
R = 0.082057
n = 0.43999
P = 1 atmosphere
T = 0 + 273 = 273
V = ???
1 * V = 0.43999 * 082057 * 273
V = 10.2 L
Answer: B

Note if you give out Brainly awards I'd sure appreciate one. This was a lot of typing

8 0

3 years ago