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2 years ago

PLEASE HELP ASAP WITH THESE THREE!

Chemistry

1 answer:

Furkat [3]2 years ago

5 0

Answer:

001: 71.2 g per ML

002: Consider a block of marble that occupies 287 cm3 and has a mass of 869 g.

a. What is its density?

d=m/v=869g/287 cm3=3.03 g/cm3

b. Will it float in:

H2O(l)? Why or why not?

Nope! Density > 1.00 g/mL

Hg(l) (density=13.55 g/mL)? Why or why not?

Yep! Density < 13.55 g/mL

003:

Use the given functions to set up and simplify

0.309 cm3

g/cm3=g/cm3

0.309cm3=g/cm3

Explanation:

http://childschemistry.weebly.com/uploads/7/8/6/3/78638416/density_key.pdf

Hope this helps I tried my best

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A gas has a pressure of 3.16 atm at STP. I have decided to transfer it to a container that is 3 times larger than the original v

Oliga [24]

Answer:

The new pressure is 1.0533 atm

Explanation:

According to Boyle's Law : The Pressure of fixed amount of gas is inversely proportional to Volume at constant temperature.

PV = Constant

P1V1 = P2V2

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P1 = 3.16 atm

Accprding to question ,

V1 = V

V2 = 3 V

Insert the value of V1 , V2 and P1 in the equation(1)

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$3.16\times V=P_{2}\times 3V$

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$3.16=P_{2}\times 3$

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2 years ago

The balanced equation for photosynthesis is shown below.

klio [65]

The true answer is: It's conserved because the total number of H atoms on each side is 12.
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2 years ago

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A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

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schepotkina [342]

Answer:

mendeleev left a space

Explanation:

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