Answer:
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Explanation:
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Answer:
Mass percent of N2H4 in original gaseous mixture = 31.13 %
Explanation:
Given:
Initial mass of gaseous mixture = 61.00 g
Initial mole of oxygen = 10.0 mol
Moles of oxygen remaining after the reaction = 4.062 mol
Moles of oxygen used = 10.0 - 4.062 = 5.938 mol
Total oxygen used in both the reactions = 10.0 parts
out of 10 parts, 3 part react with N2H4.
Now, consider the reaction of N2H4
3 moles of O2 react with 1 mole of N2H4
1.78 moles of oxygen will react with 1.78/3 = 0.5933 mol of N2H4
Molecular mass of N2H4 = 32 g/mol
Total mass = 61.0 g
The mass percent of N2H4 in the gaseous mixture can be determined through stoichiometric calculations and determining the limiting reactant. The initial and remaining amounts of O2 are used to calculate the reacted amount of O2, which then allows for the calculation of the amount of N2H4. This information is used in the mass percent formula.
The balanced reaction states that for one mole of NH3, one mole of O2 is required, while for one mole of N2H4, 3 moles of O2 are required. Thus, the initial moles of O2 were 10 moles and after reaction 4.062 moles O2 remained. Thus, the reacted amount of O2 is 10 - 4.062 = 5.938 moles. From calculating the limiting reactant and applying stoichiometry, the amount of N2H4 can be determined. We know the molar mass of N2H4 is 32 g/mole. By calculating the molar ratio, we can then calculate the mass percent of N2H4 in the mixture using the formula: (mass of N2H4 / total mass) * 100%.
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(2) methane
(3) sodium nitrate
(4) potassium chloride
Compound (3) sodium nitrate contains both ionic and covalent bonds.
Compound (3) sodium nitrate contains both ionic and covalent bonds. Sodium nitrate consists of the ions Na+ and NO3-, where the bond between Na+ and NO3- is predominantly ionic, and the bond within the NO3- ion is covalent. The sodium ion (Na+) donates an electron to the nitrate ion (NO3-), resulting in an ionic bond. However, within the nitrate ion, the nitrogen (N) and oxygen (O) atoms share electrons, forming covalent bonds.
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2.)lithium + oxygen to lithium oxide
3.)oxygen + lithium to lithium + oxide
4.)lithium oxide to lithium + oxygen
The correct answer is the second option. The wordequation that would represent the formation of lithium oxide from lithium metal and oxygen is:
lithium + oxygen to lithium oxide
Inchemical symbols, the balanced chemical equation is expressed as:
Li + O2 = Li2O
Answer : The correct option is, (2) lithium + oxygen to lithium oxide
Explanation :
When the lithium react with the oxygen to give lithium oxide as a product.
The balanced chemical reaction will be,
This reaction can be represented in words as,
Lithium react with oxygen to give lithium oxide.
or,
lithium + oxygen to lithium oxide
Hence, the correct option is, (2) lithium + oxygen to lithium oxide
A. 2PO43–(aq) + Cl– (aq) → Cl2(PO4)3(s)
B. 2Ca2+(aq) + Na+(aq) → NaCa2(s)
C. Na+(aq) + Cl– (aq) → NaCl(s)
D. 2PO43–(aq) + 3Ca2+(aq) → Ca3(PO4)2(s)
Answer:
Explanation:
In net ionic equation we remove the spectator ions. The ions which are present on both the side and are not forming any solid compound.
Let us write the balanced reaction first:
The ionic reaction is:
Thus net ionic reaction is
Carbon dioxide, ethanol, and energy
Lactic acid and energy
Glucose, oxygen, and energy
Answer:
Chlorine
Explanation:
Chlorine is a halogen that is a strong oxidizer (it takes electrons from nearby compounds). In so doing, it kills bacteria, viruses, and other microorganisms. The chlorine reacts with cell walls or other vital organic compounds (e.g., proteins) to render them useless. Chlorine is relatively inexpensive and generally easy to handle, but it is dangerous in gaseous form and highly alkaline in solution, so must be stored and handled properly.