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melamori03 [73]
3 years ago
6

The greenhouse gas released into the atmosphere as a result of livestock digesting their food is called carbon dioxide

Chemistry
1 answer:
slamgirl [31]3 years ago
3 0

Answer:

False

Explanation:

Cows and other livestock release methane, which is a gas like Carbon Dioxide, but not the same thing, even though it does contribute into Carbon Dioxide.

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Do bases lose or gain hydrogen ions?<br><br><br>Help :-)
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Acids are classified in the Bronsted Lowry acid-base theory as species, which losses hydrogen ions (H+, known as protons) in a chemical reaction (and are foundation species that accept hydrogen ions).

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What is the mass, in grams, of 6.11 mol of sulfur trioxide?
Kazeer [188]
It would be approximately 489 g, rounded for significant figures. The calculator given answer would be 489.20326. To get that, all you have to do is take your 6.11 moles and multiply it by 80.066 (the molar mass) divided by 1 mol to cancel out the unit. 
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4 years ago
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Which of the following actions will not increase the rate at which a solid dissolves in
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Answer:

O lowering the temperature of the system

5 0
3 years ago
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

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

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

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

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
4 years ago
HELP ME!! THANH YOU If an atom has 80 protons and 121 neutrons, and is not in an excited state, how many electrons would Bohr sa
wariber [46]
I think it might be 80
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3 years ago
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