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3 years ago

Compare and contrast physical layers and compositional layers.

Chemistry

1 answer:

Yuki888 [10]3 years ago

7 0

Answer:

Compositional Layers

The Earth has different compositional and mechanical layers. Compositional layers are determined by their components, while mechanical layers are determined by their physical properties. The outermost solid layer of a rocky planet or natural satellite.

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A student measured out 0.333 moles of theobromine, CyH3N402 (MM = 180.8/mol) the

Verdich [7]

Answer

2.0 x 10²³ molecules.

Explanation

Given:

The number of moles of theobromide measured out = 0.333 moles.

MM of theobromide = 180.8 g/mol

What to find:

The number of molecules of theobromide the student measured.

To go from moles to molecules, multiply the number of moles by Avogadro's number.

The Avogadro's number = 6.02 x 10²³

1 mole of theobromide contains 6.02 x 10²³ molecules.

So, 0.333 moles of theobromide measured out will have (0.333 x 6.02 x 10²³) = 2.0 x 10²³ molecules.

8 0

10 months ago

Molar mass (NH4)2SO4

Hoochie [10]

Answer:

Explanation:

Molar mass N = 14.00

Molar mass H = 1.01

Molar mass H4 = 1.01 x 4 = 4.04

Molar mass NH4 = 14.00 + 4.04 = 18.04

Molar mass (NH4)2 = 18.04 x 2 = 36.08

Molar mass S = 32.07

Molar mass O = 16.00

Molar mass O4 = 16.00 x 4 = 64.00

Molar mass SO4 = 32.07 + 64.00 = 96.07

Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>

6 0

3 years ago

A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

λ: is the decay constant = $ln(2)/t_{1/2}$

$t_{1/2}$ : is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

$N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg$

Now, at 7:00 pm we have:

$t = 7:00 pm - (5:00 pm + 90 min) = 30 min$

$N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg$

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0

2 years ago

Solve the ideal gas law equation for pressure.

posledela

Answer:

$p=\frac{nRT}{V}$

Explanation:

The ideal gas law equation is an equation that relates some of the quantities that describe a gas: pressure, volume and temperature.

The equation is:

$pV=nRT$

where

p is the pressure of the gas

V is the volume of the gas

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas (must be expressed in Kelvin)

Here we want to solve the equation isolating p, the pressure of the gas.

We can do that simply by dividing both terms by the volume, V. We find:

$p=\frac{nRT}{V}$

So, we see that:

- The pressure is directly proportional to the temperature of the gas

- The pressure is inversely proportional to the volume of the gas

6 0

3 years ago

Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc

jeka94

Answer:

$T_f$ = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

$T_i$ = 21.5°C

$C_{calorimeter}$ = 6.15 kJ/°C

$T_f$ = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate × $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

= 0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter = $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

4 0

3 years ago