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romanna [79]
2 years ago
8

How many molecules ?

Chemistry
1 answer:
dimaraw [331]2 years ago
7 0

Answer:

42 millions of single cells

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To which period does an element belong if it has its 3d orbitals filled
kotegsom [21]

fourth period

The third period is similar to the second, except the 3s and 3p sublevels are being filled. Because the 3d sublevel does not fill until after the 4s sublevel, the fourth period contains 18 elements, due to the 10 additional electrons that can be accommodated by the 3d orbitals.

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2 years ago
Where on the periodic table are the metals found?
White raven [17]
On the left side.  The alkali metals are group 1, the alkaline earth metals are group 2, and the transition metals are groups 3-12.
6 0
3 years ago
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5 0
3 years ago
How much heat is absorbed when 90.5 g of ice is heated from -11.0 °C to 145.0 °C?
Nadusha1986 [10]

Answer:

Q(total) = 283Kj

Explanation:

5 Heat Transitions …

Specific Heats => c(s) = 0.50cal/g∙⁰C,  c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C

Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g

Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.

Q = mcΔT for temperature change regions of the heating curve (single phase only)

Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)

Solid (ice) => Melting Pt  => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal

Melting (s/l) => Liquid (water) =>   Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal

Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal

Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal

Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal

Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)  

                                 = 478cal +7240cal + 9050 cal + 48,870cal + 2036cal

                                 = 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj

4 0
4 years ago
How many grams are in 9.97 moles of Be(NO3)2?<br> Use two digits past the decimal for all values.
Simora [160]

Answer:

1,869.97 grams of Be(NO3)2

Explanation:

Be(NO3)2 = Be  N2  O6

Be=9.012182g/mole

N2=28.0134g/mole

O6=96g/mole

therefore Be(NO3)2 gives you 187.56g in one mole

so 9.97 moles means there is 9.97 times more

9.97mole Be(NO3)2 * 187.56g Be(NO3)2/1mole Be(NO3)2 = 1,869.97g of Be(NO3)2

4 0
3 years ago
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