write the name of the following compounds NaCI ,SO3 ,MgCI, N2H4 and LiF. thank you!
1. write the name of the following compounds NaCI ,SO3 ,MgCI, N2H4 and LiF. thank you!
Answer:
sodium chloride,sulfur trioxide,magnesium chloride,hydrazine,lithium fluoride
2. what is the empirical formula of the compound hydrazine with a molecular formula N2H4?
Answer:
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3. 7. N2H4 (hydrazine)what is the atoms of N2H4
Answer:
hydrogen bat hydrazine
the atoms ofN2H4 IS HCFIoz
4. A. Analyze and classify the following chemical formulas written below as to empirical formula or molecular formula.1.)CH=_________2.)HO=_________3.)C6H12O6=________4.)N204=__________5.)N2H4=__________6.)C2H2=__________7.)H2O2=__________8.)C2H1206=_________9.)NO2=__________10.)NH2=___________PLEASE ANSWER ༼;´༎ຶ ༎ຶ༽
yan pa sa pic po thank you
lab ya
Answer:
????
hope it helps .
5. ACTIVITY: identify the elements in each chemical formula and give the number of atoms examples:H2O(water)=2 H and 10NaC(salt)=1Na and 1Cl_______1.NaCN(sodium cyanide)_______2.H2S(hydrogen sulphide)_______3.C2H6(ethane)_______4.SnF2(stannous flouride)_______5.CO(carbon manoxide)_______6.C3H8O3(glycerol)_______7.N2H4(hydrazine)i skip num.8_______9.C3H8O(isopropyl alcohol)_______10.C4H10(butane)pa help po ako >.<
Answer:
Sory hindi ako magaling dyan
Answer:
Pasenya po nakalimutan ko na to kung paano last year pa yon kasi
6. 2 NH3(g) + 1 half O2(g) → N2H4(g) + H2O(l) calculate the value of ΔH
Explanation:
To calculate the value of ΔH (the enthalpy change) for the given reaction:
2 NH3(g) + 1/2 O2(g) → N2H4(g) + H2O(l)
We need to know the enthalpy of formation of each species involved in the reaction, as well as the stoichiometric coefficients of the reactants and products.
Using standard enthalpies of formation, we have:
ΔH = [ΔHf(N2H4) + ΔHf(H2O)] - [2ΔHf(NH3) + 1/2 ΔHf(O2)]
Where ΔHf represents the enthalpy of formation.
The values for the enthalpies of formation are:
ΔHf(N2H4) = 50.6 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
ΔHf(NH3) = -46.1 kJ/mol
ΔHf(O2) = 0 kJ/mol
Substituting these values into the equation, we get:
ΔH = [(50.6 kJ/mol) + (-285.8 kJ/mol)] - [2(-46.1 kJ/mol) + 1/2(0 kJ/mol)]
ΔH = -235.6 kJ/mol
Therefore, the value of ΔH for the given reaction is -235.6 kJ/mol. This indicates that the reaction is exothermic, meaning that it releases energy in the form of heat.
7. does n2h4 satisfy the octet rule?
Explanation:
does n2h4 satisfy the octet rule
the answer is yes.
8. What is the mass of 0.443 mol of hydrazine, N2H4?
Answer: 32.0452 g/mol
14.18486 gramsExplanation:
#carryonlearning9. What is the mass of 0.443 mol of hydrazine, N2H4? (show your solution)
Answer:
14.1960 grams
Explanation:
Molar mas of N2H4 is 32.04516 grams
so 1 mol = 32.04516
given mol is 0.443
so 0.443 × 32.04516 = 14.1960 grams
Sana Makatulong
Goodluck
Patama pag Mali
10. Hi guys just want to know if it is a spontaneous or non spontaneous a. 2H2O(g)⇌2H2(g)+O2(g) with ΔG = +228.59 kJmol b. N2H4(l)+O2(g)→N2(g)+2H2O(l) with ΔH∘ =-1244.4 kJmol
Answer:
A.spontaneous
B.spontaneous
Explanation:
PA BRAINLIEST THANK YOU :)
11. calculate the formula mass of n2h4 with solution plss
Answer:
32.0452 g/mol
Formula: N2H4
Density: 1 g/cm³
Boiling point: 114 °C
12. Put the following in descending order of mass? 1.0 mole N2H4 2.0 mole N2 3.0 mole NH3 25.0 mole H2
Explanation:[tex]33 = 33333333 \times 44 + 5739724 \: \: \: \: \: [/tex]
13. Find the ∆H for the reaction below, given the following reactions and subsequent ∆H values: N2H4 (l) + H2 (g) → 2NH3 1. N2H4 (l) + CH4O (l) → CH2O (g) + N2 (g) + 3H2 (g) ∆H = 37 kJ 2. N2 (g) + 3H2 (g) → 2NH3 (g) ∆H = -46 kJ 3. CH4O (l) → CH2O (g) + H2 (g) ∆H = -65 kJ
Answer:
The ∆H for the reaction N2H4 (l) + H2 (g) → 2NH3 (g) is -9 kJ. This can be calculated by adding the ∆H values for the individual reactions:
∆H = (1) + (2) + (3)
∆H = 37 kJ -46 kJ - 65 kJ
∆H = -9 kJ
14. Activity 4.Name the following covalent compounds.1. SO22. NO23. CO24. H205. NF36. N2H47. N2058. NBr39. As2O410. CCL4
Answer:
4 is water
Explanation:
sorry that's what i know
Answer:Sulfate
Nitrate
Carbon dioxide
Water
Di ko po alam
Explanation:
15. How many particles of Cl2 is needed to produce 32g of N2H4?
Answer:
0.867 g.
Explanation:
http://www.csun.edu/~hcchm003/100/supp5soln.pdf
16. What is the mass of 0.443 mol of hydrazine,N2H4
Answer:
1 moles N2H4 to grams = 32.04516 grams.
Answer:
14.2 g
Explanation:
I hope it's help #CARRYONLEARNING
17. LSSON1. Calculate the mass of dinitrogen tetroxide, N2O4 that is added to 50 g of dinitrogen tetrahydride,N2H4 in this chemical reaction:2 N2H4(1) + N204(1)→ 3 N2(g) + 4H2O(1)
[tex]\tt{\huge{\red{Solution:}}}[/tex]
Based on the balanced chemical equation, 1 mole of N₂O₄ is stoichiometrically equivalent to 2 moles of N₂H₄. Note that the molar masses of N₂O₄ and N₂H₄ are 92.011 g/mol and 32.0452 g/mol, respectively.
[tex]\begin{aligned} \text{mass of} \: \text{N}_2\text{O}_4 & = 50 \: \cancel{\text{g} \: \text{N}_2\text{H}_4} \times \frac{1 \: \cancel{\text{mol} \: \text{N}_2\text{H}_4}}{32.0452 \: \cancel{\text{g} \: \text{N}_2\text{H}_4}} \times \frac{1 \: \cancel{\text{mol} \: \text{N}_2\text{O}_4}}{2 \: \cancel{\text{mol} \: \text{N}_2\text{H}_4}} \times \frac{\text{92.011 g} \: \text{N}_2\text{O}_4}{1 \: \cancel{\text{mol} \: \text{N}_2\text{O}_4}} \\ & = \boxed{\text{72 g} \: \text{N}_2\text{O}_4} \end{aligned}[/tex]
Hence, 72 g of N₂O₄ is added to 50 g of N₂H₄.
[tex]\\[/tex]
Note: Kindly swipe the screen to the left to see the continuation of the answers on the right side.[tex]\\[/tex]
#CarryOnLearning
18. What is the mass of 0.55 mol of hydrazine, N2H4?PAHELP POOO
Answer:32.04516Explanation:converting between grams Hydrazine and mole. You can view more details on each measurement unit: molecular weight of Hydrazine or mol The molecular formula for Hydrazine is N2H4.
19. Hydrazine, N2H4(l), is used in rocket fuels. The thermochemical equation for the combustion of hydrazine is N2H4(l)+O2(g) → N2(g)+2H2O(l) ∆H= -622.4 kj. What quantity of heat is liberated by the combustion of 1.0 g of N2H4(l)?Show the solutions:
Answer:
anong grade ka na ba anong klaseng ano toh comment mo na lang
20. If 145g N2H, and 175g N20, are allowed to react, 2 N2H4+ N2O4 → 3 N2 + 4 H20, determine the (a) limitingreactant; (b) mass (in grams) of excess reactant; and (c) mass (in grams) of water.
a. N₂O₄ as a limiting reactant
b. mass of excess reactant(N₂H₄) that reacts := 122.41 g
c. mass of water : 137.52 g
Further explanationA reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Limiting reactants are reactants that limit the outcome of the reaction. Where if two substances are reacted there is a possibility that both substances are used up or one is used up and the other is left.
The method that can be used to find limiting reactants :
divide the number of moles of known substances by their respective coefficients, and the small quotient will be used up or as a limiting reactant
Reaction
2N₂H₄ + N₂O₄ → 3N₂ + 4H₂O
mol N₂H₄ :
= 145 g : 32,0452 g/mol
= 4.52
mol N₂O₄ :
= 175 g : 92.011 g/mol
= 1.91
mol : coefficient :
N₂H₄ : N₂O₄ = 4.52/2 : 1.91/1 = 2.26 : 1.91
N₂O₄ as a limiting reactant
mass of excess reactant(N₂H₄) that reacts :
= 2/1 x mol N₂O₄ x 32,0452 g/mol
= 2/1 x 1.91 x 32,0452 g/mol
= 122.41 g
the unreacted mass of N₂H₄:
= 145 - 122.41
= 22.59 g
mass of water :
= 4/1 x mol N₂O₄ x 18 g/mol
= 4/1 x 1.91 x 18 g/mol
= 137.52 g
Learn more
the mole fractions of the solute and the solvent
https://brainly.ph/question/2062141
moles of CaCO3
https://brainly.ph/question/2356002
the mole ratio of Al to Cl2
https://brainly.ph/question/2124499
#LetsStudy
21. One mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (There is no change in the oxidation state of hydrogen). a–1 b–3 c+3 d+5
Explanation:
that all the nitrogen a4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the newppears in the new compound, what is the oxicompound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (Thedation state of nitrogen in Y? (There is no change in the oxidation state of hydrogen).
