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ololo11 [35]
3 years ago
10

Which produces a higher than normal tide? a Spring Tide or Neal Tide? and why

Chemistry
1 answer:
Mila [183]3 years ago
7 0

Answer:

spring tide

Explanation:

when both the sun and moon aligned th affect of each is added together producing higher than normal tides called SPRING TIDES

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If 83.6 grams of H2 and 257 grams of N2 react, how many grams of ammonia will be produced?
SSSSS [86.1K]

The mass of ammonia that would be produced is 312.5 g

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

3H₂(g) + N₂(g) → 2NH₃(g)

This means

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to produce 2 moles of ammonia.

First, we will determine the number of moles of each reactant present

For Hydrogen (H₂)

Mass = 83.6 g

Molar mass = 2.016 g/mol

Using the formula

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Number of moles of H₂ present = \frac{83.6}{2.016}

∴ Number of moles of H₂ present = 41.468254 moles

For Nitrogen (N₂)

Mass = 257 grams

Molar mass = 28.0134 g/mol

∴ Number of moles of N₂ present = \frac{257}{28.0134}

Number of moles of N₂ present = 9.174181 moles

Since,

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to produce 2 moles of ammonia

Then,

27.522543 moles of hydrogen gas will react with the 9.174181 moles of nitrogen gas to produce 18.348362 moles of ammonia

∴ 18.348362 moles of ammonia will be produced during the reaction

Now, for the mass of ammonia that would be produced

From the formula

Mass = Number of moles × Molar mass

Molar mass of ammonia = 17.031 g/mol

Mass of ammonia that would be produced = 18.348362 × 17.031

Mass of ammonia that would be produced = 312.49095 g

Mass of ammonia that would be produced ≅ 312.5 g

Hence, the mass of ammonia that would be produced is 312.5 g

Learn more here: brainly.com/question/13902065

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CF4 is the compound that has polar bonds, but dipole moment of O. C-f bond is a polar bond which is covalent. It has dipole of 4 polar C.F bonds and results in the overall monopolar molecule.
CF4 it has no net dipole moment.
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the periodic table created by Dmitri Mendeleev is very useful to that contain a large amount of information about an atom of a p
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13. Copper(II) chloride and silver acetate Acid/Base:
Lubov Fominskaja [6]

Answer:

13. CuCl_2 \hspace{0.1cm}+ CH_3COOAg -> AgCl \hspace{0.1cm} + (CH_3COO)_2Cu

This is a double displacement reaction, not an acid-base reaction

14.HF\hspace{0.1cm}+Ba(OH)_2 ->\hspace{0.1cm}H_2O\hspace{0.1cm}+BaF_2

Ba(OH)_2 -> Base

HF -> Acid

15. HCl\hspace{0.1cm}+Mg(OH)_2->\hspace{0.1cm}H_2O\hspace{0.1cm}+MgCl_2

MgCl_2 -> Base

HCl -> Acid

16.CH_3COOH\hspace{0.1cm}+LiOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+CH_3COOLi

LiOH -> Base

CH_3COOH -> Acid

17.H_2SO_4\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaHSO_4

NaOH -> Base

H_2SO_4-> Acid

18. H_2SO_4\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaHSO_4

NaOH -> Base

H_2SO_4-> Acid

19.H_2SO_4\hspace{0.1cm}+NaHCO3->\hspace{0.1cm}H_2O\hspace{0.1cm}+CO_2+\hspace{0.1cm}Na_2SO_4

NaHCO_3-> Base

H_2SO_4-> Acid

20.H_3PO_4\hspace{0.1cm}+Zn(OH)_4->\hspace{0.1cm}H_2O\hspace{0.1cm}+Zn_3(PO_4)_4

Zn(OH)_4> Base

H_3PO_4-> Acid

21. HF\hspace{0.1cm}+Na_2SO_3->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaF\hspace{0.1cm}+\hspace{0.1cm}SO_2

NaSO_2> Base

HF-> Acid

22.HCl\hspace{0.1cm}+Na_2SO_3->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaCl\hspace{0.1cm}+\hspace{0.1cm}SO_2

NaSO_2> Base

HCl-> Acid

23.HCl\hspace{0.1cm}+Na_2S->\hspace{0.1cm}H_2S\hspace{0.1cm}+NaCl

Na_2S> Base

HCl-> Acid

24. NH_4Cl\hspace{0.1cm}+NaOH->\hspace{0.1cm}H_2O\hspace{0.1cm}+NaCl+{0.1cm}NH_3

NaOH> Base

NH_4Cl-> Acid

Explanation:

All acid-base reaction have the same products, H_2O and a salt. Whit this in mind the acids would the compounds that produces the hydronium ion (H^+) and the bases would be the compounds that produces the hydroxide ion (OH^-)

8 0
3 years ago
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