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Kryger [21]
3 years ago
15

Chemical properties of sodium

Chemistry
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

Density 0.97 g.cm -3 at 20 °C

Melting point 97.5 °C

Boiling point 883 °C

oxidation states +1, −1 (rare)

Explanation:

Have a nice day

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Why do scientists look for patterns in the world?
PSYCHO15rus [73]

Answer:

B. Patterns can help explain observations.

Explanation:

Hope this helps

6 0
3 years ago
At 1 atm, how much energy is required to heat 75.0 g H 2 O ( s ) at − 20.0 ∘ C to H 2 O ( g ) at 119.0 ∘ C?
Hitman42 [59]

Answer:

238,485 Joules

Explanation:

The amount of energy required is a summation of heat of fusion, capacity and vaporization.

Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)

m (mass of water) = 75 g

Lf (specific latent heat of fusion of water) = 336 J/g

C (specific heat capacity of water) = 4.2 J/g°C

∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

Lv (specific latent heat of vaporization of water) = 2,260 J/g

Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J

3 0
3 years ago
what separation procedure is most likely to happen? .a. freezing a compound will break it down into elements .b. filtering a mix
SOVA2 [1]

Explanation:

i think its B but i'm not fully sure (Correct)

8 0
3 years ago
C. Another radioisotope of americium exists which has an atomic mass of 242.
ollegr [7]

Answer:

1 g  

Explanation:

The half-life of Am-242 (16 h) is the time it takes for half of it to disappear.

We can make a table of the mass left after each half-life.

\begin{array}{cccc}\textbf{No. of} & & \textbf{Percent} & \textbf{Mass}\\\textbf{Half-lives} & \textbf{Time/h} & \textbf{Remaining} & \textbf{Remaining/g}\\0& 0 & 100 & 8\\1 & 16 &50 & 4\\2 & 32 & 25 & 2\\3 & 48 & 12.5 & 1\\4 & 64 & 6.25 & 0.5\\\end{array}

The mass remaining after 48 h  is 1 g.

7 0
3 years ago
Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)
steposvetlana [31]
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 
6 0
3 years ago
Read 2 more answers
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