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3 years ago

A location along stream where most erosions occur

Chemistry

1 answer:

vagabundo [1.1K]3 years ago

4 0

Answer:

At a stream's headwaters, often high in the mountains, gradients are steep. The stream moves fast and does lots of work eroding the stream bed.As a stream moves into lower areas, the gradient is not as steep. Now the stream does more work eroding the edges of its banks.

hope that helps bby<3

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The atmosphere surrounding Earth helps to maintain the various climates found around the world and keeps Earth from becoming ext

riadik2000 [5.3K]

Answer:

The atmosphere traps heat energy from the Sun and energy radiated from Earth's surface helping to maintain Earth's climate

Explanation:

Earth's atmosphere keeps much of the Sun's energy from escaping into space. This process, called the greenhouse effect, keeps the planet warm enough for life to exist. The atmosphere allows about half of the Sun's heat energy (50%) to reach Earth's surface.

4 0

3 years ago

If chemical bonds didn’t

hram777 [196]

All of the above answers are true right answer is option D

because, no bond no molecule no interaction no life,

no bond= only element will exist

Energy would not exist coz energy consumption and excretion takes place during bond formation and bond breaking process!

3 0

3 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

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

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

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

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

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

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

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 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?

Bogdan [553]

The second option only.

LiOH, Li₂SO₄.

<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

Ca(OH)₂ and
LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

$\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}$ .

The formula for calcium sulfate $\text{CaSO}_4$ in option A is spelled incorrectly. Why? The charge on each calcium $\text{Ca}^{2+}$ is +2. The charge on each sulfate ion ${\text{SO}_4}^{2-}$ is -2. Unlike $\text{Li}^{+}$ ions, it takes only one $\text{Ca}^{2+}$ ion to balance the charge on each ${\text{SO}_4}^{2-}$ ion. As a result, $\text{Ca}^{2+}$ and ${\text{SO}_4}^{2-}$ ions in calcium sulfate exist on a 1:1 ratio.

$2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}$ .

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

$\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}$ .

The ion ${\text{NH}_4}^{+}$ acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

$2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}$ .

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0

3 years ago

Why is drinking urine unhealthy?<br> whats in it?

Kryger [21]

This is by far probably the best question I've answered all day.
It's not unhealthy to drink it.
Urine consists of 95% water and 5% nutrients

7 0

3 years ago

Read 2 more answers