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3 years ago

How did the Columbus storm start

Chemistry

1 answer:

Anna71 [15]3 years ago

3 0

Answer:

The Columbus Day Storm of 1962 (also known as the Big Blow,[2] and originally as Typhoon Freda) was a Pacific Northwest windstorm that struck the West Coast of Canada and the Pacific Northwest coast of the United States on October 12, 1962. It is considered the benchmark of extratropical wind storms. The storm ranks among the most intense to strike the region since at least 1948, likely since the January 9, 1880 "Great Gale" and snowstorm. The storm is a contender for the title of most powerful extratropical cyclone recorded in the U.S. in the 20th century; with respect to wind velocity, it is unmatched by the March 1993 "Storm of the Century" and the "1991 Halloween Nor'easter" ("The Perfect Storm"). The system brought strong winds to the Pacific Northwest and southwest Canada, and was linked to 46 fatalities in the northwest and Northern California resulting from heavy rains and mudslides.

Explanation:

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Elements that want to gain one outer electron?

mario62 [17]

Answer:

halogens

Explanation:

Elements that are nonmetals tend to gain electrons and become negatively charged ions called anions

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3 years ago

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Which two atoms would typically form a covalent bond?

MrMuchimi

Answer:

two atoms both non metals

Explanation:

because if the bond is formed between atoms with similar electronegativity since covalent bonding is the sharing of electrons

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4 years ago

Which phases can coexist at equilibrium at point A on the phase diagram?

AURORKA [14]

Answer:

<u>A. solid and gas</u>

Explanation:

It is called sublimation point

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Nitrogen monoxide and hydrogen react to form nitrogen and water, like this: 2NO+2 H2(g)-N29)+2H2O(9) Write the pressure equilibr

lara31 [8.8K]

Answer:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp

Explanation:

The reaction is:

2NO + 2H₂ → N₂ + 2H₂O

The expression for Kp (pressure equilibrium constant) would be:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²

There is another expression for Kp, where you work with Kc (equilibrium constant)

Kp = Kc (R.T)^Δn

where R is the Ideal Gases constant

T° is absolute temperature

Δn = moles of gases formed - moles of gases, I had initially

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4 years ago

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene is burned in a bomb calorim

Serggg [28]

Answer:

$\Delta E_{rxn}$ for combustion of naphthalene is -5164 kJ/mol

Explanation:

$\Delta E_{rxn}=C_{calorimeter}\times \Delta T_{calorimeter}$

where, C refers heat capacity and $\Delta T$ refers change in temperature.

Here, $\Delta T_{calorimeter}=(32.33-24.25)^{0}\textrm{C}=8.08^{0}\textrm{C}$

So, $\Delta E_{rxn}=(5.11\frac{kJ}{^{0}\textrm{C}})\times (8.08^{0}\textrm{C})=41.3kJ$

$\Delta E_{rxn}$ is generally expressed in terms of per mole unit of reactant. Also, $\Delta E_{rxn}$ should be negative as it is an exothermic reaction (temperature increases).

Molar mass of naphthalene is 128.17 g/mol

So, 1.025 g of naphthalene = $\frac{1.025}{128.17}moles$ of naphthalene

= 0.007997 moles of naphthalene

$\Delta E_{rxn}=-\frac{41.3kJ}{0.007997mol}=-5164kJ/mol$

5 0

3 years ago