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3 years ago

What is a disadvantage to hydropower dams?

Chemistry

1 answer:

Sindrei [870]3 years ago

8 0

B because it does happens when you make a dam waters is upstream and when the dam breaks it flows down stream

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Write the balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI.

Kaylis [27]

If it is just a simple double-replacement reaction, then the balanced equation is simply:

Pb(ClO3)2(aq) + 2 NaI(aq) = PbI2 + 2 Na(ClO3)

However, if this is a precipitation reaction, we must identify which compound will precipitate (is insoluble), either NaClO3 or PbI2

NaClO3 is soluble, PbI2 is not. Only the lead cations and iodine anions will participate in the reaction.

thus, the balanced equation is: Pb^(2+) + 2I^(-) ==> PbI2

4 0

3 years ago

What do scientists use to determine what the interior of a moon is like?

polet [3.4K]

Use sound waves, and examine samples of earth, examine meteorites

7 0

3 years ago

What does ozone do to make life on earth possible?

forsale [732]

Answer - Ozone layer helps keep the harmful ultra violet rays hitting and keeping the atmospheric pressure in.

Reasoning - Without the ozone layer the whole planet would have a loss of water and especially people burning very easily from the sun. <span />

6 0

3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

3 0

3 years ago

If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make

vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

2H₂ + O₂ -----------> 2H₂O

Now look at the reaction for mole ratio

2H₂ + O₂ -----------> 2H₂O

2 mole 2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

2H₂ + O₂ -----------> 2H₂O

2 mole (2 g/mol) 2 mole (18 g/mol)

4 g 36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

4 g of H₂ ≅ 36 g of H₂O

30 g of H₂ ≅ X of H₂O

Do cross multiplication

X of H₂O = 30 g x 36 g / 4 g

X of H₂O = 270 g

Now convert grams of H₂O into moles

No. of moles = mass in grams/molar mass

Put values in above formula

No. of moles = 270 g / 18 (g/mol)

No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

5 0

3 years ago