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3 years ago

How do glaciers cause erosion and deposition

2 answers:

7 0

when glaciers (slowly) glide down a mountain they carry down earth and rock and brings it down. After the glacier has completely melted the earth from the mountain id deposited to the lower ground.

koban [17]3 years ago

6 0

Like flowing water, flowing ice erodes the land and deposits the material elsewhere. Glaciers cause erosion in two main ways: plucking and abrasion.
Plucking is the process by which rocks and other sediments are picked up by a glacier. They freeze to the bottom of the glacier and are carried away by the flowing ice.
The other way you'll need more informations because my Knowledges are limited by here.

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most planets have --- but none has a --- like earth does. A. a biosphere; hydrosphere., B. an atmosphere; cryosphere., C. an atm

liraira [26]

C. is the answer. i know because i had this question not that long ago.

5 0

3 years ago

Read 2 more answers

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

SnO2 + 2 H2 ——> Sn + 2 H2O

SpyIntel [72]

Answer:

0.15g

Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

Mass of SnO₂ = ?

Solution:

To solve this problem, we have to work from the known to the unknown specie;

SnO₂ + 2H₂ → Sn + 2H₂O

Ensure that the equation given is balanced;

Now,

the known species is water;

6.02 x 10²³ molecules of water = 1 mole

1.2 x 10²¹ molecules of water = $\frac{1.2 x 10^{21} }{6.02 x 10^{23} }$ = 0.2 x 10⁻²moles

Number of moles of water = 0.002moles

From the balanced chemical equation:

2 mole of water is produced from 1 mole of SnO₂

0.002 moles of water will be produced from $\frac{0.002}{2}$ = 0.001moles

To find the mass;

Mass = number of moles x molar mass

Molar mass of SnO₂ = 118.7 + 2(16) = 150.7g/mol

Mass = 0.001 x 150.7 = 0.15g

3 0

3 years ago

Which type of changes are melting, freezing, and boiling?

tatiyna

Answer:

Adding heat.

Decreasing and increasing temperatures.

Explanation:

4 0

3 years ago

Read 2 more answers

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0

3 years ago