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weqwewe [10]
4 years ago
7

Oxbows form in rivers because the water erodes soil from the outside of a curve and deposits material at the inside. The river i

n the picture has an oxbow bend.
If the river is left alone, how might the river appear 50 years from now?
Erosion by the river will make the oxbow straighten out.
The river will most likely dry up.
The oxbow may become isolated and become a lake.
The river will most likely look exactly the same.
Chemistry
2 answers:
LenaWriter [7]4 years ago
7 0
<span>The oxbow may become isolated and become a lake.

</span>
nikklg [1K]4 years ago
6 0

The correct answer is

The oxbow may become isolated and become a lake.

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Does anybody know this ?
geniusboy [140]

Explanation:

density =mass/volume

density = 603/281

density = 2.1459 g/cm³

3 0
3 years ago
Which compounds are classified as Arrhenius acids?
Olenka [21]

Answer:The correct answer is option 4.

Explanation:

Arrhenius acids are those compounds which gives H^+ ions when dissolved in their aqueous solution.

HA(aq)\rightarrow H^++A^-

Arrhenius bases are those compounds which gives OH^- ions when dissolved in their aqueous solution.

BOH(aq)\rightarrow OH^-+B^+

HBr \& H_2SO_4 are Arrhenius acids because they form H^+ions in their respective aqueous solution.

HBr(aq)\rightarrow H^++Br^-

H_2SO_4(aq)\rightarrow 2H^++SO_{4}^{2-}

Hence, the correct answer is option 4.

6 0
4 years ago
Read 2 more answers
(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak
Taya2010 [7]

Answer:

Here's what I get  

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

                   H₂ +    I₂    ⇌ 2HI

I/mol·L⁻¹:    0.30   0.15         x

2. Calculate the concentration of HI

Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} =  5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

 

7 0
3 years ago
Some type of path is necessary to join both half-cells in order for electron flow to occur.
klemol [59]

The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

Explanation:

Flow of electrons is possible with the help of a conducting medium like metal wire.

A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.

If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.

Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.

4 0
3 years ago
For the reaction: h2(g) + cl2(g) → 2hcl(g), how many moles hcl will be produced from 10.0 g of h2? the reaction occurs in the pr
motikmotik
Hope this would help you

3 0
3 years ago
Read 2 more answers
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