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Kaylis [27]
3 years ago
7

How much pollution is in urban streams ?

Chemistry
1 answer:
Marizza181 [45]3 years ago
7 0

Explanation: Pollution in urban streams (commonly caused by improperly disposed waste and/or acid rain) always has a negative effect on the environment. <em>Urban rivers are extremely polluted in some areas</em>, such as India. Actions have been taken to clear the streams, and progress has been made.

Hope this helps!

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What is the approximate temperature of the descending air at an elevation of 1,600 meters?
Margarita [4]
According to Diagram B, look at the 1600 elevation until you see the descending air line touches it. Then look down at the temperature at the bottom of the graph. It is between 0 degrees to 5 degrees.

The only number that is between that range is 2 degrees C.
3 0
3 years ago
One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO
zhenek [66]
Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s)  and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
                                                                                     = 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
                                                         = 1000 x 10³ g / 226 g mol⁻¹
                                                         = 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x \frac{3}{2}
                                                     = 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
                                        = 703540.02 g
                                        = 703.540 kg

Hence, the theoretical yield of Na₂CO₃ =  703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
                            = 92.34%
7 0
3 years ago
What do you put in the calculator to get the molar mass of Mg3P2
olga55 [171]

Input the atomic masses of Mg and P to give 134.84g/mol

Explanation:

The molar mass of a substance (atom or molecule or compound) is the mass in grams of one mole of the substance:

When dealing with an element the molar mass is the relative atomic mass expressed as g/mol.

For compounds, you add the atomic masses of the component atoms and you sum up.

You simply input the atomic mass of 3 atoms of Mg and 2 atoms of P

Atomic mass of Mg = 24.3g/mol

                             P = 30.97g/mole

Molar mass of Mg₃P₂ = 3(24.3) + 2(30.97) = 134.84g/mol

learn more:

Molar mass brainly.com/question/2861244

#learnwithbrainly

6 0
3 years ago
Find the ratio of the effusion rate of hydrogen gas to the effusion rate of krypton gas.
Zielflug [23.3K]

This problem could be solved through the Graham’s law of effusion (also known as law of diffusion). This law states that the ratio of the effusion rate of the first gas and effusion rate of the second gas is equivalent to the square root of the ratio of its molar mass. Thus the answer would be 0.1098. 

6 0
3 years ago
Read 2 more answers
Acetic acid and water react to from hydronium cation and acetate anion, like this: HCH3CO2 (aq) + H2O (I) → H3O+(aq) + Ch3CO2-(a
Roman55 [17]

Answer:

1) Greater than zero, and equal to the rate of the reverse reaction

2) Greater than zero, but less than the rate of the reverse reaction

3) Greater than zero, and equal to the rate of the reverse reaction

Explanation:

A reaction system is said to be in equilibrium when the rate of forward reaction is equal to the rate of reverse reaction.

Before we remove HCH3CO2 from the system, the system was in equilibrium. Recall that when a system is in equilibrium, the rate of forward reaction is equal to the rate of reverse reaction. The rate of reaction is greater than zero because products are being formed as the reactants interact with each other.

When HCH3CO2 is removed from the system, the equilibrium position shifts towards the left hand side hence the rate of reverse reaction is greater than the rate of forward reaction.

When the system attains equilibrium again, the rates of forward and reverse reaction become equal.

8 0
2 years ago
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