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Dmitriy789 [7]
3 years ago
13

Two processes are described below: Process A: rapid moving water strikes a rock on the bottom of a stream and breaks it into pie

ces Process B: glaciers moving along a rocky surface form cracks in the rocks
which statement is true about the processes?

both represent physical weathering

both represent chemical weathering

Process a is physical weathering in process B is chemical weathering

process a is chemical weathering and process B is physical weathering
Chemistry
2 answers:
worty [1.4K]3 years ago
7 0

Answer:

both represent physical  weathering

Explanation:

maksim [4K]3 years ago
4 0
Both represent physical weathering
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You perform a single extraction with 100 mL of chloroform to remove compound A from 100 mL of an aqueous solution. The partition
Mice21 [21]

Answer:

Percentage of mass of solute in aqueous phase is 28.57%.

Explanation:

(q)_{aq}- Fraction of solute in aqueous phase.

(q)_{aq}=\frac{V_{aqueous}}{DV_{organic}+V_{aqueous})}

V_{aqueous} = Used volume of aqueous solution

V_{organic} = Used volume of organic solution

D = partition coefficient.

(q)_{aq}=\frac{100ml}{(2.5\times100)+100\,ml}=0.2857

Mass of solute in aqueous phase = 0.2857\times 5g=1.428g

Mass\,percentage = \frac{1.428}{5.0}\times 100=28.57

Therefore, Percentage of mass of solute in aqueous phase is 28.57%.

5 0
3 years ago
The car has a mass of 1200kg. The car has a forward force of 3400N but friction has a force of 400N going against it. What is th
jonny [76]

The acceleration of the car : 2.5 m/s²

<h3>Further explanation</h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  

∑F = m. a  

<h3 />

Mass of the car = 1200 kg

Forward force = 3400 N

Friction force = 400 N

The direction of motion of the friction force will be opposite to the forward force, so that both of them can be subtracted to get the value of the net force (net force: combination / sum of all forces acting on an object)

\tt \sum F=3400-400=3000~N

So the acceleration :

\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{3000}{1200}\\\\a=2.5~m/s^2

6 0
3 years ago
The normal boiling point of a substance occurs when its vapor pressure is 1.00 atm, which means that at this temperature, the li
Sladkaya [172]

Answer:

See explanation below

Explanation:

For start, we need some values here to do this exercise.

In general, you can calculate the normal boiling point of any substance by using the Clausius Clapeyron equation which is the following:

ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)

Where:

P1 and P2: pressure of the substance at T1 and T2.

ΔHvap: enthalpy of vaporization of the substance. In the case of bromine is 29.6 kJ/mol

R: constant gas. In this case is 8.3145 J/mol K

T1 and T2: temperature of the substance.

In order to calculate the normal boiling point, we will assign that value to T2, and the pressure would be 1 atm or 1.01x10^5 Pa

T1 and P1 would be temperature and pressure of this substance at any condition. For this example, I will take the fact that Bromine has 22000 Pa at 20 °C (or 293.15 K)

With this data, let's replace in the clausius Clapeyron equation:

ln(1.01x10^5 / 22000) = -29600/8.3145 (1/T2 - 1/293.15)

-1.5241 * 8.3145 / 29600 = (1/T2 - 1/293.15)

-4.281x10^-4 + 1/293.15 = 1/T2

T2 = 1 / 2.9831x10^-3

T2 = 335.22 or 62.07 °C

The real one is 59 °C so, the difference in the result may come with the values of P1 and T1 that may be not accurate.

7 0
3 years ago
Last year Frank had a total income of $58,800. He sold a house and made a profit of $27,940. He also had monthly income of $80 f
ipn [44]

Answer: D.) 0.90.

Explanation:

3 0
2 years ago
A 25.888 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 73.464 g of wa
brilliants [131]

Answer:

Weight % of NH₃ in the aqueous waste  = 2.001 %

Explanation:

The chemical equation for the reaction

\\\\NH_3} + HCl -----> NH_4Cl

Moles of HCl = Molarity × Volume

= 0.1039 × 31.89 mL × \frac{1 \  L}{1000 \ mL}

= 0.0033 mole

Total mass of original sample = 25.888 g + 73.464 g

= 99.352 g

Total HCl taken for assay = \frac{10.762 \ g}{99.352 \ g}

= 0.1083 g

Moles of NH₃ = \frac{0.0033 \  mol}{0.1083}

= 0.03047 moles

Mass of NH₃ = number of  moles × molar mass

Mass of NH₃ = 0.03047 moles × 17 g

Mass of NH₃  = 0.51799

Weight % of NH₃  = \frac{0.51799 \ g}{25.888 \ g} * 100%%

Weight % of NH₃ in the aqueous waste  = 2.001 %

4 0
4 years ago
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