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2 years ago

Part A Summarize your research, and describe the attributes of the savanna.

2 answers:

Nutka1998 [239]2 years ago

8 0

Savannas are grasslands found in Africa, Australia, South America, and India. The climate is warm or hot. Savannas experience both dry and rainy seasons. Each year, savannas receive about 20 to 50 inches of rain in a span of six to eight months. The other months are dryer. There’s also a threat of wildfires because of the severe dryness during these months. Grasslands are true to their name because their vegetation is mostly grass. Savannas are home to a wide range of animals, including different species of beetles and insects, which many creatures consume. Squirrels, mice, elephants, gazelles, and antelopes are the plant-eating animals that live in the savanna. Hyenas, lions, and leopards are some of the meat-eating animals in these areas. Some of the nonliving components of the system include very porous soil that allows water to drain easily.

zhuklara [117]2 years ago

7 0

Answer:

Savanna regions have two distinct seasons - a wet season and a dry season. There is very little rain in the dry season. In the wet season vegetation grows, including lush green grasses and wooded areas.

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If Log 4 (x) = 12, then log 2 (x / 4) is equal to

Alexus [3.1K]

The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

= log₂(2²⁴/2²)

= log₂(2²⁴ ⁻ ²)

= log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

#SPJ9

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1 year ago

This reaction appears to be destroying the logs

Alex_Xolod [135]

What exactly is the question you are asking?

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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.

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