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3 years ago

Which horizon has the least weathered rocks? How do you know?

Chemistry

1 answer:

ollegr [7]3 years ago

4 0

Answer:

C: The C horizon is a subsurface horizon. It is the least weathered horizon. Also known as the saprolite, it is unconsolidated, loose parent material. The master horizons may be followed by a subscript to make further distinctions between differences within one master horizon.

Explanation:

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What is the charge on every atom? why is this the charge

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Protons, neutron, and elecrons

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3 years ago

For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the

Aleks04 [339]

Answer:

1.2x10⁻⁵M = Concentration of the product released

Explanation:

Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:

A = E*b*C

Where A is the absotbance of the solution: 0.216

E is the extinction coefficient = 18000M⁻¹cm⁻¹

b is patelength = 1cm

C is concentration of the solution

Replacing:

0.216 = 18000M⁻¹cm⁻¹*1cm*C

<h3>1.2x10⁻⁵M = Concentration of the product released</h3>

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3 years ago

The diagram shows an electric generator connected to a lightbulb. Not all of

BaLLatris [955]

Answer:

The lightbulb transforms it into thermal energy.

Explanation:

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3 years ago

Explain, in terms of electrons, why the compound containing calcium and chlorine is classified as an ionic compound.

Pavel [41]

Because the reaction occurs between a metal and non metal
CL=2,8,7
C

4 0

3 years ago

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:

Margarita [4]

Answer:

108.6 g

Explanation:

2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the PV=nRT formula to calculate the number of nitrogen moles:

P = 1.00 atm

V = 56.0 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 0 °C ⇒ 0 + 273.2 = 273.2 K

Inputting the data:

1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K

n = 2.5 mol

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

2.5 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.67 mol NaN₃

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

1.67 mol * 65 g/mol = 108.6 g

6 0

3 years ago