If you have 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, how much ammonia (in grams) is produced? 3H2 + N2 --> 2NH3
Answers
If we have 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, 39.66g ammonia is produced.
What is mole concept?
Avogadro's number is the number of units in one mole of any substance and equals to 6.02214076 × 10²³. The units can be electrons, atoms, ions, or molecules.
No. of moles is defined as a particular no. of particles that we can calculate with the help of Avogadro’s number.
Mass of a particular product is also find out by stoichiometry of a reaction as per the no. of mole given in the reaction.
Given,
3H₂ + N₂ --> 2NH₃
3.50 moles of hydrogen
5.00 moles of nitrogen
Limiting reagent for the reaction is H₂.
3 moles of H₂ produce 2 moles NH₃
1 moles of H₂ produce 2/3 moles NH₃.
3.5 moles of H₂ produce (2÷3)×3.5 moles NH₃ = 2.33 moles
1 mole of NH₃ = 17g of NH₃
2.33 mole of NH₃ = 39.66g of NH₃
Therefore, If we have 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, 39.66g ammonia is produced.
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3H2 + N2 --> 2NH3
3.5 moles H2 (2 moles NH3/3 moles H2) (17 grams NH3/1 mole NH3) = 40 grams NH3
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Answers
The equivalent of 750 joules is 0.750 kJ.
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Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent.
So we get:
So you get about:
So, the percent composition of Hydrogen in NH4HCO3 is 6.3%
Hope this Helps! :D
i hope this helps hydrogen in a compound =5 h atoms are present
therefore % of composition =5/79 *100 so it equals 6.3%
Answers
A controlled experiment is an experimental setup designed to test the hypotheses. It has one or more conditions (independent variables) and measures (dependent variables). It is necessary to change more than one variable, but all os the experimental conditions will be controlled so that only the variables being examined change and the amount or way of change is measured.
- When [S] << Km, the reaction is second order and V0 depends on [S] and [Et].
- Their kcat is a second order rate constant.
- The lower their Km, the better they recognize their substrate, but the lower their reaction rate.
- When [S] << Km, V0 depends on [S] and [Et].
Answers
Answer:
1. True. 2. True. 3. Not true. 4. True. 5. True
Explanation:
1. Yes, because if the amount of substrate i much greater than of competitive inhibitor then the probability of substrate to bind to ferment is much higher than of inhibitor (if we have noncompetitive inhibitor it damages the structure of active site and the substrate concentration does not have a role in reaction rate).
2. Yeah, because then the michaelis-menten equation will transform into [tex} V0=(kcat*[E]*[S])/Km [/tex] and it is a second order equation.
3. No, because it is measured in sec-1 and that means it is 1 rate constant.
4. True, if the lower Km the better is binding and due to that rate is slower because it's harder for substrate to unbind.
5. The same as question two.
Answers
.... I hope it helps
Final answer:
Two oxygen atoms form a stable molecule by sharing two pairs of electrons, forming a double covalent bond. This is an important process for achieving the required octet electron configuration for both oxygen atoms.
Explanation:
When two oxygen atoms combine to form a molecule, the type of bond they form is a double covalent bond. An atom of oxygen has six electrons in its valence shell and it requires two more to achieve stability, hence, the oxygen atom achieves this equilibrium by sharing two pairs of electrons forming a double covalent bond. This is similar to the example of hydrogens and an oxygen atom sharing electrons to form a water molecule, where covalent bonds are also present.
The creation of a double bond occurs as these atoms share more than one pair of electrons to attain the required octet configuration. This could also be seen in bonds formed between carbon and oxygen atoms in formaldehyde (CH₂O), or between carbon atoms in ethene (C₂H4).
The covalent bond formed between these two oxygen atoms is, therefore, a key component for them to form a stable molecule.
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Answers
Answer:
Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3
Explanation:
Lead nitrate is Pb(NO3)2
Sodium Chloride is NaCl
The basic reaction is:
Pb(NO3)2 + NaCl = PbCl2 + NaNO3
Now we need to balance the equation. Note that the lead nitrate contributes 2 (NO3) units, but there is only 1 NO3 on the product side. So start by adding a second NaNO3:
Pb(NO3)2 + NaCl = PbCl2 + 2NaNO3
Now we have 2 Na atoms on the right side, so add a "2" to the NaCl:
Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3
This is starting to look good. Lets check the atom/unit counts on both sides of the equation:
Reactants Products
Pb 1 1
NO3 2 2
Na 2 2
Cl 2 2
The equation is now balanced. Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3