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ahrayia [7]
3 years ago
10

Which is the last thing to be seen before a total eclipse?

Chemistry
1 answer:
kupik [55]3 years ago
8 0
<span>When the Moon has nearly blocked out the sun at the last few minutes before the total solar eclipse, you will see some points of light surrounding the edges of the dark Moon. These look like a string of beads wrapped around the edges and are appropriately called Baily's Beads. They are named after the astronomer Francis Baily who first noticed them in the 18th century. These beads only appear for a few brief minutes before the total solar eclipse occurs.</span>
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Four nails have a total mass of 4.42 grams how many moles of iron atoms do they contain
deff fn [24]

Answer:4.42 g= 1 mol/55.845 =.079 moles of Fe

Explanation:Given 4.42 grams of Fe. The atomic weight of Fe(iron) found on the periodic table is 55.845. Divide grams by the atomic weight to convert to moles.

5 0
4 years ago
Draw the structure of the following atoms<br> (1) 19/9 F<br> (2)28/14 SI
Svet_ta [14]

Answer:

might help

Explanation:

8 0
3 years ago
There are 2 gasses, A, B. They weigh 2.46g and 0.5g respectively, and the Volume of A is 3 times the volume of B. A has a molecu
dangina [55]

Answer:

B

Explanation:

molecular mass of B is 28

8 0
2 years ago
Read 2 more answers
A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. what is the percent yield for this rea
Masja [62]

CaCO3(s) ⟶ CaO(s)+CO2(s) 

<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>

<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3, therefore 0.0131 moles CO2 should also be formed. 
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>

Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>

4 0
3 years ago
Read 2 more answers
How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?
vodomira [7]

Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

Water ⇒ C = 1 cal/g°C

m = 65.8 g

Ti = 31.5°C

Tf = 36.9°C

Heat, Q = ?

2) Formula:

Q = mCΔT

3) Calculations:

Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal

4) You can convert from calories to Joules using the conversion factor:

1 cal = 4.18 J

⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J

3 0
3 years ago
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