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1 year ago

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Choose all the answers that apply. Lakes are recharged by _____. 1.evaporation 2.precipitation 3.rivers 4.transpiration 5.ground

water seepage 6.melting snow and ice

2 answers:

sweet-ann [11.9K]1 year ago

6 0

2. Precipitation, because when it rains the water that was taken when it was evaporated is replaced and it fills the lake back up

julsineya [31]1 year ago

6 0

THE ANSWER IS PRECIPITATION. PRECIPITATION IS RAIN. EVERY TIME IT RAINS, EVERY BODY OF WATER IS INVOLVED IN THIS PROCESS. THE OCEAN, LAKES, PONDS, ETC

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Fill in the blanks to determine the number of protons and electrons in an oxygen ion. (consult the periodic table to determine t

saveliy_v [14]

The protons in an nucleus of an atom will not change unless a nuclear reaction takes place. The number of protons is equal to the atomic number of the element. For a neutral atom, the number of electrons and protons are equal. When they are unequal, then the atom occurs as an ion. It will has a net charge with it. The ion O²⁻ has a net charge of negative positive 2 because it has 2 more electrons than its protons. Since neutral oxygen has 8 protons, then O²⁻ ion has 8 protons and 10 electrons.

5 0

2 years ago

Read 2 more answers

We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find

iragen [17]

Explanation:

a) Using the provided information about the density of gold, the sample size, thickness, and the following equations and comersion factors, find the area of the gold leaf:

$V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V$

Gold $_{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}$

$1 \mu=10^{-6} \mathrm{m}$

First, find the volume of the sample and then find the area of the sample.

$V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3}$ $=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}$

$V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}$

b. Using the provided information from part $a$ ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

$V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}$

$h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}$

8 0

2 years ago

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 <em> initial partial pressure</em>

-x -x +x +x <em> react</em>

105.1-x 7.96-x x x <em>equilibrium</em>

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

Read 2 more answers

The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C

Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $460^oC$ = $5.8\times 10^{-6}s^{-1}$

$K_2$ = rate constant at $T_2$ = $4\times K_1$

$Ea$ = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $460^oC=273+460=733K$

$T_2$ = final temperature = ?

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

$\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]$

$T_2=791.1K$

Therefore, the final temperature would be, 791.1 K

4 0

2 years ago

Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

asambeis [7]

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

6 0

1 year ago