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Svetllana [295]
3 years ago
12

Why is the temperature needed to freeze ocean water lower than the temperature needed to freeze the surface of a freshwater lake

? Explain in detail!
Answer ASAP pls
I will give brainliest
Chemistry
1 answer:
shepuryov [24]3 years ago
6 0
Ocean water freezes just like freshwater, but at Lower temperature. Fresh water freezes At 32°F but see water freezes at about 28.4°F because of the salt in it it can be melted down to use as drinking water
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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
I bet no one can solve this
Nikitich [7]
The answer is 89.4 g/mol cuz the two units carry positives so that means 8.50 flows for 2 more hours than usual.
3 0
3 years ago
Which mixture is classified as a solution?
svlad2 [7]
Salt water is considered to be a solution
5 0
3 years ago
How many electrons would an atom need to have before it can begin filling the 3s sub level?
schepotkina [342]
Well electrons are like electricity well it is partly if it went 3 sub levels the electrons would blow up some electricty<span />
5 0
3 years ago
Use the following balanced reaction to solve 1-3:
tino4ka555 [31]

Answer:

1.96g H2

Explanation:

P4 = 123.9g/mol

H2 = 2.02g/mol

\frac{20g P_{4} }{1} * \frac{1 mol P_{4} }{123.9g} * \frac{6 mol H_{2} }{1 mol P_{4} } * \frac{2.02g H_{2} }{1 mol H_{2} } = 1.96g H2

6 0
3 years ago
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