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My name is Ann [436]
3 years ago
14

Janelle hikes along the beach every morning at the same time. Throughout the month, she notices the shoreline follows a pattern

of receding and advancing. What causes the shoreline to follow this pattern?
tides


water cycle


chemical weathering


erosion and deposition
Chemistry
2 answers:
avanturin [10]3 years ago
8 0

Answer:

Tides

Explanation:

just did it:)

Zina [86]3 years ago
7 0

Answer:

Erosion and deposition

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Antarctica, almost completely covered in ice, has an area
Marysya12 [62]

<u>Answer:</u> The mass of ice is 2.39\times 10^{22}g

<u>Explanation:</u>

We are given:

Area of Antarctica = 5,500,000mi^2=5,500,000\times 2.59\times 10^{10}=142.45\times 10^{15}cm^2      (Conversion factor:  1mi^2=2.59\times 10^{10}cm^2  )

Height of Antarctica with ice = 7500 ft.

Height of Antarctica without ice = 1500 ft.

Height of ice = 7500 - 1500 = 6000 ft = 182.88\times 10^3cm     (Conversion factor:  1 ft = 30.48 cm)

To calculate mass of ice, we use the equation:

\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

We are given:

Density of ice = 0.917g/cm^3

Volume of ice = Area × Height of ice = 142.45\times 10^{15}cm^2\times 182.88\times 10^3cm=26051.26\times 10^{18}cm^3

Putting values in above equation, we get:

0.917g/cm^3=\frac{\text{Mass of ice}}{26051.26\times 10^{18}cm^3}\\\\\text{Mass of ice}=(0.917g/cm^3\times 26051.26\times 10^{18}cm^3=2.39\times 10^{22}g

Hence, the mass of ice is 2.39\times 10^{22}g

5 0
3 years ago
a = Proton transfer d = SN2 Nucleophilic substitution g = Nucleophilic subs at carbonyl(acyl Xfer) b = Lewis acid/base e= Electr
marissa [1.9K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution to this question is shown on the second uploaded image

Explanation:

4 0
3 years ago
A process at constant T and P can be described as spontaneous if ΔG &lt; 0 and nonspontaneous if ΔG &gt; 0. Over what range of t
creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

                              2 SO3(g)   →          2 SO2(g) + O2(g)

ΔHf° (kJ mol-1)  -395.7                        -296.8

S° (J K-1 mol-1)  256.8                         248.2              205.1

ΔH° =  kJ

S° =  J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine  ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where  ν  is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS =  ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) =  2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ =  197.8 - T(0.1879)

and the reaction will become spontaneous when the term  T(0.1879)  becomes greater that 197.8,

0 = 197.8 - 0.1879 T  ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0
3 years ago
Which statement best explains why ice cubes will melt in a glass of tea?
Radda [10]

Answer:

The ice cubes gain energy

Explanation:

The ice cubes are gaining energy from the water's carbon dioxide, or H2O. They then melt because they cannot hold that energy.

5 0
3 years ago
You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2
avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).  

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0
3 years ago
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