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solong [7]
3 years ago
5

Olive oil has a density of 0.92 g/mL. How much would 1 liter of olive oil weigh in grams?

Chemistry
1 answer:
svp [43]3 years ago
4 0
Dimension analysis is to be used to solve this problem. First convert 1L to milliliters. That is equivalent to 1000 ml. Then by dimension analysis, multiply the volume ( 1000ml) to the density of oil (0.92 g/ml) resulting to the answer: 920 grams. 
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How many moles are in 4177g of aluminum chloride
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The mole<span> is the </span>unit of measurement<span> in the </span>International System of Units<span> (SI) for </span>amount of substance<span>. It is defined as the </span>amount<span> of a </span>chemical substance<span> that contains as many representative particles, e.g., </span>atoms<span>, </span>molecules<span>, </span>ions<span>, </span>electrons<span>, or </span>photons<span>, as there are atoms in 12 </span>grams<span> of </span>carbon-12<span> (</span>12<span>C), the </span>isotope<span> of </span>carbon<span> with </span>relative atomic mass<span> 12 by definition. 
so to solve the moles, divide the mass with molar mass

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3 years ago
What is the dominant greenhouse component and why?
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Answer: HOPE THIS HELPED! :D

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Explanation:

7 0
3 years ago
Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2
PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

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

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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