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2 years ago

How is the weathering of rocks by water different from the weathering of rocks by ice?

Chemistry

1 answer:

ehidna [41]2 years ago

6 0

Answer:

The formation of ice can also break rocks. If water gets into a crack in a rock and then freezes, it expands and pushes the crack further apart. When the ice melts later, water can get further into the crack. When the water freezes, it expands and makes the crack even bigger.

Explanation:

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Molecular compounds are capable of forming multiple covalent bonds. the compound below that contains a triple bond is:

alexira [117]

The answer is <u>CO(g).</u>

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1 year ago

A 3.00 L flexible container holds a sample of hydrogen gas at 153 kPa. If the pressure increases to 203 kPa and the temperature

dybincka [34]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 V1 / P2

V2 = 153 x 3.00 / 203

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2 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2

Marizza181 [45]

Answer:

$\Delta _rH=-1124.14kJ/mol$

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

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

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

$\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}$

In such a way, by using the NIST database, we find that:

$\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol$

Thus, we plug in the enthalpies of formation to obtain:

$\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol$

Best regards!

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2 years ago

Ok this is so weird but like how do you expose fake friends for talking about you? I know they did it but i dont know how to tal

murzikaleks [220]

Answer:

just do it

Explanation:

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3 years ago

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