Why are all guys the same. They all cheat and break my heart. Why do they do it?

Chemistry

2 answers:

Paul [167]3 years ago

5 0

Some do it because that is the way they are.

Some times they do it because in reality they do not care

And some do it just to be mean

But there is someone out there for you.

It is going to take time to find the right guy.

Because not all guys are like this

PilotLPTM [1.2K]3 years ago

3 0

Not all guys will break your heart and act selfish. When you find the right one he will treat you like a queen and you’ll never have to worry about him being unfaithful. I know it doesn’t help a hurting heart, but you WILL find your soulmate one day! Just keep faith! Best wishes!

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0.0145 moles of helium gas are introduced into a balloon so that the volume of the balloon is 2.54 liters. An additional amount

olga_2 [115]

Answer:

4.43L is final volume of the ballon

Explanation:

Avogadro's law of ideal gases states that equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

The formula is:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

Where V and n are volume and moles of the gas in initial and final conditions.

If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

$\frac{2.54L}{0.0145mol} =\frac{V_2}{0.0253mol}$

V₂ = 4.43L is final volume of the ballon

6 0

2 years ago

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$