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Illusion [34]
4 years ago
10

Which is a chemical property of soda ash?

Chemistry
1 answer:
Anvisha [2.4K]4 years ago
4 0
Dry, white powder, soluble in water to form a slightly basic solution
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Which is NOT part of the definition of matter
qaws [65]

Answer:

Everything that has mass and takes up space is matter. Every day, you find something usual that either does't have mass or else don't take up space. Those things are non-matter. Basically, any type of energy or any abstract concept is an example of something that does not have matter.

Explanation:

7 0
3 years ago
Read 2 more answers
Calculate the equilibrium constant for the decomposition of water 2h2o(l)  2h2(g) + o2(g) at 25°c, given that g°f (h2o(l)) = –
kow [346]

Answer:

2.6 ×10^-42

Explanation:

From

∆G= -RTlnK

∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1

R= 8.314 Jmol-1K-1

T= 25°C + 273= 298K

-237.2×10^3= 8.314 × 298 × ln K

ln K= -237.2×10^3/2477.572

K = 2.6 ×10^-42

3 0
4 years ago
Complexes containing metals with d10 electron configurations are typically colorless because ________. Complexes containing meta
algol [13]

Answer:

there is no d electron that can be promoted via the absorption of visible light

Explanation:

One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.

The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.

For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.

Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.

8 0
3 years ago
Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 , has a solubility constant of Ksp = 2.34 × 10 − 59 , and dissociates according to Ca
GenaCL600 [577]

Answer: 4M

Explanation:

5 0
3 years ago
if a sample of gas at 25.2 c has a volume of 536mL at 637 torr, what will its volume be if the pressure is increased to 712 torr
nignag [31]
Considering ideal gas:
PV= RTn

T= 25.2°C = 298.2 K

P1= 637 torr = 0.8382 atm

V1= 536 mL = 0.536 L

:. R=0.082 atm.L/K.mol

:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))

:. n= O.0184 mol

Then,
P2= 712 torr = 0.936842 atm

V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)

:.V2 = 0.4796 L
OR
V2 = 479.6 ml

6 0
3 years ago
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