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AVprozaik [17]
3 years ago
10

O2 + 2H2 --> 2H2O in this equation, the O2 is a

Chemistry
1 answer:
mart [117]3 years ago
7 0

Answer:

this is too easy just do it yourself

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Which formula represents an organic compound?cah2c4h8h2o2p2o5?
Dmitry_Shevchenko [17]
Answer is: C₄H₈.
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CaH</span>₂ (calcium hydride) is alkaline earth hydride, inorganic compound.
H₂O₂(hydrogen peroxide) is pale blue, clear, inorganic <span>liquid.
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3 0
3 years ago
Draw the product formed when the structure shown below undergoes solvolysis in ch3ch2oh with heat.
mars1129 [50]

structure is attached below

7 0
3 years ago
An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density
V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

Density = \dfrac{4 \times M_w}{N_A \times a^3}

where

M_w = molar mass of the compound

N_A= avogadro's constant

a^3 = the volume of a unit cell

Given that:

Density (\rho) = 19.30 g/cm³

a = 0.408 nm

a = 0.408 \times 10^{-9} \times 10^{2} \ cm

a = 4.08 \times 10^ {-8} \ cm

∴

19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}

M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}

M_w=197.37 \ g/mol

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0
3 years ago
Question 22
Mrrafil [7]
HELPPPPPPPPPPPPPPPPPPPPPPPf it is a good idea to
3 0
3 years ago
The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
5 0
3 years ago
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