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3 years ago

Among the elements magnesium, sodium and silicon, which hås the highest second ionisation

Chemistry

1 answer:

FinnZ [79.3K]3 years ago

3 0

Answer:

Explanation:

silicon has highest ionization energy ionization energy increases from left to right in a period because electron are added to same shell and increase in added electron shield each other poorly from the nuclear attraction

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According to the theory of plate tectonics,

11Alexandr11 [23.1K]

A I’m sure of it because it only makes since one would think ✊

3 0

3 years ago

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

5 0

3 years ago

Oxygen freezes at -219°c and boils at -183°c in which state is oxygen at -200°c

vova2212 [387]

oxygen is in gaseous state at this temperature

4 0

3 years ago

Which mineral group dose copper belong

RideAnS [48]

Answer:

Native Elements; Metallic Elements

Explanation:

7 0

3 years ago

If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees

telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁ + q₂ = 0

n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃

∴ n = 0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g

q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J

q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0

ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol

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3 years ago