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3 years ago

5

in 4 season countries, the temperatire may drop till below 0°c during the peak of winter. why does the surface of the river or l

ake freeze?

1 answer:

nikitadnepr [17]3 years ago

4 0

Answer:

Explained below

Explanation:

Normally, when the temperature of water gets colder than 4° C, it starts expanding and will become less dense the colder it becomes. As the temperature gets closer to freezing point, the colder water will float to the top and the warmer water sinks to the bottom. This is mainly because of how the density of water behaves with falling temperatures.

Thus, the lake/river water will freeze from top to down as a a result of that phenomenon of colder water floating to the top and the warmer water sinking to the bottom.

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A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH₄NO₃ and 4.42 g of (NH₄)₃ PO₄ in enoug

Alexandra [31]

In finding the molarity of a solution, we use the following formula:

$M=moles soluteL solution$

What is Molarity?

The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.

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The molar mass of NH4NO3 and (NH4)3PO4 are 80.043 g/mol and 149.0867 g/mol, respectively.

$molesNH+4inNH4NO3=5.66 g80.043 g/mol×1molNH+41molNH4NO3=0.0707 mol$

$molesNH+4in(NH4)3PO4=4.42 g149.0867 g/mol×3molNH+41mol(NH4)3PO4=0.0889 mol$

$total molesNH+4=0.0707 mol+0.0889 mol=0.1596 mol$

$molesPO3−4in(NH4)3PO4=4.42 g149.0867 g/mol×1molPO3−41mol(NH4)3PO4=0.0296 mol$

$[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4$

$[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4$

Therefore, $[PO3−4]\\$ has a molarity of $1.48×10−3 M PO3−$

To learn more about Molarity click on the link below:

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1 year ago

Using the Lewis dot structures of magnesium and oxygen, predict the ionic formula

Blizzard [7]

Answer:

(edit: nvm I figured it out, here is the answer)

Explanation:

7 0

3 years ago

A gas is initially at a pressure of 4.7 atm and a volume of 7.5 L. When the volume is increased to 10.5 L, what is the new press

Roman55 [17]

Answer:

P2=3.36 atm

Explanation:

P1×V1=P2×V2

P1×V1÷V2=P2×V2÷V2

P2=4.7×7.5÷10.5

P2=3.36 atm

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2 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

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