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poizon [28]
3 years ago
6

Hydrogenous sediments originate from elements in?

Chemistry
1 answer:
natka813 [3]3 years ago
5 0
seawater - formed by the precipitation of minerals from the ocean's water. :)
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What are the Sun's layers, beginning with the innermost and moving out.
Lemur [1.5K]
Core, radiative zone, convective zone, photosphere, chromosphere, corona 
4 0
3 years ago
Propane can be turned into hydrogen by the two-step reforming process. In the first step, propane and water react to form carbon
JulsSmile [24]

Answer:

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

Explanation:

Propane can be turned into hydrogen by the two-step reforming process.

In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:

C₃H₈(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g)

In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:

CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)

In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.

C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

4 0
3 years ago
When bacteria reproduces asexually are the offspring uniform or diverse?
mario62 [17]

Answer:

Reproduction may be asexual when one individual produces genetically identical offspring, or sexual when the genetic material from two individuals is combined to produce genetically diverse offspring.

4 0
3 years ago
Read 2 more answers
Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -> Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which
kotykmax [81]

Answer:

Balanced reaction: 3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O

S is oxidized and N is reduced.

NO_{3}^{-} is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO

\Rightarrow Oxidation: ZnS\rightarrow Zn^{2+}+S

Balance charge: ZnS-2e^{-}\rightarrow Zn^{2+}+S  ...............(1)

\Rightarrow Reduction: NO_{3}^{-}\rightarrow NO

Balance H and O in acidic medium : NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O

Balance charge: NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O ...............(2)

[3\timesEquation-(1)] + [ 2\timesEquation-(2)]:

3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of NO_{3}^{-} to NO. Therefore N is reduced.

NO_{3}^{-} consumes electron from ZnS. Therefore NO_{3}^{-} is the oxidizing agent and ZnS is the reducing agent.

7 0
4 years ago
Uranium-232 has a half-life of 68.8 years. After 344.0 years, how much uranium-232 will remain from a 100.0-g sample?
amm1812

Answer:  3.13 g

Explanation:

Radioactive decay follows first order kinetics.

Half-life of uranium-232 = 68.8 years

\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{68.8}= 0.010072674 year^{-1}

N=N_o\times e^{-\lambda t}

N = amount left after time t

N_0 = initial amount

\lambda = rate constant

t= time

N_0 = 100 g, t= 344 years, \lambda=0.010072674 years^{-1}

N=100\times e^{- 0.010072674 years^{-1}\times 344 years}

N=3.13g


8 0
3 years ago
Read 2 more answers
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