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goldfiish [28.3K]
3 years ago
15

Which area is MOST likely to support a marsh ecosystem? a sunny, open area with lots of sand dunes a shallow, sheltered area nea

r an estuary a rocky, coastal area with very high tides deep ocean water miles from a river's mouth
Chemistry
2 answers:
Vedmedyk [2.9K]3 years ago
7 0

Answer:

coastal area with very high tides deep ocean water miles from a river's mouth

Explanation:

Marshes are a type of wetland ecosystem where water covers the ground for long periods of time. Marshes are dominated by herbaceous plants, such as grasses, reeds, and sedges(National Geographic Society).

Animals found in a marsh include;  frogs, toads, turtles, snakes, mammals, birds and insects. A marsh is a wetland that is dominated by herbaceous rather than woody plant species(Wikipedia). It is characterized by very high tides deep ocean water miles from a river's mouth.

irga5000 [103]3 years ago
6 0

Answer:

i also need help with this question

Explanation:

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Calculate the pH of a 0.50 M HIO. The Ka of hypoiodic acid, HIO, is 2.3x10–11.0.305.325.479.474.80
never [62]

Answer:

pH = 5.47

Explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka = \frac{[H+][IO-]}{[HIO]} = 2.3 * 10⁻¹¹

At equilibrium:

  • [HIO] = 0.5 M - x
  • [H⁺] = x
  • [IO⁻] = x

<u>Replacing those values in the equation for Ka and solving for x:</u>

Ka=\frac{x^2}{0.5-x}=2.3*10^{-11} \\x^2=(2.3*10^{-11})(0.5-x)\\x^2=1.15*10^{-11}-2.3*10^{-11}x\\x^2+2.3*10^{-11}x-1.15*10^{-11}=0\\x=3.39*10^{-6}

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47

7 0
3 years ago
If chlorine gas is bubbles through an aqueous solution of sodium iodide,the result is elemental iodine and aqueous sodium chlori
Mrac [35]
The balanced reaction that describes the reaction of chlorine gas and sodium iodide to produce elemental iodine and sodium chloride in aqueous solution is expressed Cl2+2NaI= I2 + 2NaCl. This kind of reaction is called single replacement reaction where the anion, in this case, is only replaced
3 0
3 years ago
A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

                                   = 0.528 g

  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

brainly.com/question/1603500

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5 0
1 year ago
THE RIGHT ANSWER WILL RECIEVE A BRAINLEST AND POINTS!!!
earnstyle [38]
If you drop a bath bomb into water, then it will fizz because a chemical reaction is taking place.
8 0
3 years ago
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A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,
lara31 [8.8K]

Answer:

The answer to your question is:   T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L                    V2 = 2.78 L

P1 = 2.40 atm               P2 = 1.97 atm

T1 = 325°K                    T2 = ?

Formula

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Process

            T2 = (P2V2T1) / (P1V1)

            T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

            T2 = 1779.895 / 7.56

            T2 = 235.44 °K

4 0
3 years ago
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