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Anon25 [30]
3 years ago
12

The standard gibbs free energy of formation of ___ is zero: 1. h2o(l) 2. na(s) 3. h2 (g)

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
3 0

3. H2 (g) has a standard gibbs free energy of 0

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How many bonding pairs in br2o
Evgen [1.6K]
There are 22 bonding parts
5 0
3 years ago
Calculate the amount of heat required to raise the temperature of a 24 g sample of water from 9°C to 23°C.
borishaifa [10]

Answer:

1400KJ/mol⁻¹

Explanation:

Amount of heat required can be found by:

Q = m × c × ΔT

<em>Where m is the mass, c is the specific heat capacity (4.2KJ for water) and ΔT is the change in temperature.</em>

Q = 24 × 4.2 × (23 - 9)

= 24 × 4.2 × 14

=   1411.2KJ/mol⁻¹

= <u>1400KJ/mol⁻¹</u>  (to 2 significant figures)

7 0
3 years ago
Identify a reason that chemical reactions release energy during the reaction process.
Bess [88]

Answer:

option b

Explanation:

When the energy is released the process is called exothermic reaction. This happens when the bonds are broken in the reactants and the system release energy.

5 0
3 years ago
Read 2 more answers
Help
Jet001 [13]

Answer:

a

Explanation:

The formation of ion occurs when an atom that is said to be neutral gains or losses electrons.

At the time it gains electrons, it is regarded that a negative ion (anion) is formed.

When it loses electron, it is regarded that a positive ion (cation) is formed.

Atomic number = No of protons and electrons occurring in a neutral atom.

Given that:

Protons = 14

electron = 18

Net Charge = no of proton - no of electron

= 14 - 18

= -4

Mass number = 14 + 15 = 29

So, the chemical symbol = Si^{-4}

For ion with

27 proton, 32 neutrons and 25 electrons

Net charge = 27 - 25

= +2

Mass number = 27 + 32 = 59

Thus, the chemical symbol = Co^{+2}

6 0
3 years ago
For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con
Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

A\rightarrow B

Expt    [A] M        [B] M         Rate [M/s]

1          3.40         4.16           1.82*10^-4

2         4.59         4.16           3.32*10^-4

3.        3.40          5.46          1.82*10^-4

Rate = k[A]^{x}[B]^{y}

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2

To find y:

Divide rate of expt 3 by expt 1

\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0

Therefore: x = 2, y = 0

Rate = k[A]^{2}[B]^{0}

To find k

Use rate for expt 1:

k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1

6 0
3 years ago
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