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2 years ago

which event would be less widely visible from Earth: a partial lunar eclipse or a total lunar eclipse? explain please

Chemistry

1 answer:

adell [148]2 years ago

6 0

Answer:

I had the same question and I put a total lunar eclipse.

Explanation:

A total lunar eclipse would be less widely visible because, in a partial lunar eclipse the moon only has to partly be in the Earth's shadow.

I don't know if this is helpful or not, but this is what I put if you still needed it.

Still stuck? Get 1-on-1 help from an expert tutor now.

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The intersection angle of a 3 degree curve is 45.2 degrees. What is the length of the curve? of Select one: O a. 455 m O b.573 m

ValentinkaMS [17]

Explanation:

Relation between length of a curve and angle is as follows.

l = $R \times \Theta$

where, R = radius of curve

$\Theta$ = angle in radians

Also, l = $R \times \Theta \times \frac{\pi}{180}$ .......... (1)

If curve has a degree of curvature $D_{a}$ for standard length s, then

R = $\frac{s}{D_{a}} \times \frac{180}{\pi}$ ........... (2)

Now, substitute the value of R from equation (2) into equation (1) as follows.

l = $\frac{s \times \Theta}{D_{a}}$

If s = 30 m, then calculate the value of l as follows.

l = $\frac{s \times \Theta}{D_{a}}$

= $30 \times \frac{45.2}{3}$

= 452 m

thus, we can conclude that the length of the curve is 452 m.

7 0

3 years ago

Dovator [93]

Answer:

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

3 0

3 years ago

1. The solubility of AgNO3 at 20°C is 222.0g AgNO3/100g H2O. What mass of AgNO3 can be dissolved in 250 g of water at 20°C? Reca

neonofarm [45]

Answer :

(1) The mass of silver nitrate is, 555 g

(2) The solubility of the gas will be, 0.433 g/L

<u>Solution for Part 1 :</u>

From the given data we conclude that

In 100 gram of water, the amount of silver nitrate = 222 g

In 250 gram of water, the amount of silver nitrate = $\frac{222}{100}\times 250=555g$

Therefore, the mass of silver nitrate is, 555 g

<u>Solution for Part 2 :</u>

Formula used : $S_1P_1=S_2P_2$ (at constant temperature)

where,

$S_1$ = initial solubility of methane gas = 0.026 g/L

$S_2$ = final solubility of methane gas

$P_1$ = initial pressure of methane gas = 1 atm

$P_2$ = final pressure of methane gas = 0.06 atm

Now put all the given values in the above formula, we get the solubility of methane gas.

$(0.026g/L)\times (1atm)=S_2\times (0.06atm)$

$S_2=0.433g/L$

Therefore, the solubility of the gas will be, 0.433 g/L

5 0

3 years ago

What is the name of the polyatomic in MgSO4?

blsea [12.9K]

Answer:

Magnesium Sulfate

Explanation:

4 0

3 years ago

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The energy released by nuclear fission can be used to ________ and turn a turbine to produce ________ energy.

enyata [817]

Heat water; mechanical (the movement of a turbine is based off of mechanical energy, not chemical or potential).

6 0

3 years ago

Read 2 more answers