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3 years ago

When is the next lunar Eclipse and how do they happen?

Chemistry

1 answer:

kondaur [170]3 years ago

6 0

°°The next Lunar Eclipse is on March 23, a Wednesday THIS YEAR - 65 days left!

°° A Lunar Eclipse happens when the moon passes directly behind the earth into its umbra, or, shadow. This can occur only when the sun, Earth and moon are aligned exactly, or very closely so, with the Earth in the middle. Hence, a Lunar Eclipse can occur only the night of a full moon. The type and length of an eclipse depends upon the Moon's location relative to its orbital nodes. Isn't that cool?

↑ ↑ ↑ Hope this helps! :D

You might be interested in

Calculate the molar concentration of the Br⁻ ions in 0.51 M MgBr2(aq), assuming that the dissolved substance dissociates complet

Y_Kistochka [10]

MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
MgBr2(s) --> Mg+(aq) + 2 Br⁻(aq)
[Br⁻] = 0.51 mol MgBr2/1L × 2 mol Br⁻ / 1 mol MgBr2 = 1.0 M
The answer to this question is [Br⁻] = 1.0 M

6 0

2 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

brainly.com/question/2082986?referrer=searchResults

brainly.com/question/1657608?referrer=searchResults

I hope it helps you!

6 0

2 years ago

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.

6 0

3 years ago

a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses

Sergio039 [100]

Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

$PV=nRT$

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

$P=n\\or\\\frac{P}{n}=constant$

$\frac{P_1}{n_1}=\frac{P_2}{n_2}$

Initial conditions:

$P_1=93kPa\\n_1=5\text{ puffs}$

Final conditions: When additional 3 puffs of air is added

$P_2=?kPa\\n_2=8\text{ puffs}$

Putting the values, in above equation, we get

$\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa$

3 0

2 years ago

Water has a high specific heat because: Select one: a. it is a poor insulator b. hydrogen bonds must be broken to raise its temp

Crank

<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>

Explanation:

Water has high specific heat due to hydrogen bonds present in it.
The Ionisation of water does not affect the specific heat of the water.
On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
On increasing the temperature, there is the breaking of bonds hence option (b) is correct.

3 0

3 years ago