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2 years ago

Water air multiple cells a place to live

Chemistry

2 answers:

Illusion [34]2 years ago

6 0

Answer:

I guess it WATER

Dmitriy789 [7]2 years ago

5 0

It’s most likely missing water
plants need to be watered in order to survive

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If a sodium ion has 11 protons, 12 neutrons, and 11 electrons, what is the atomic mass of the atom?

iragen [17]

Answer:

Explanation:

we do not care about electrons, so 11 + 12 = 23

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3 years ago

What would happen to the annual rates of limestone (CaCO3) precipitation in the ocean if the ocean water were to become much war

stiv31 [10]

Answer:

CaCO3 exoskeleton dissolves in acidic water

Explanation:

The increasing CO2 level makes the ocean water acidic and hence reduces the pH. In such acidic environment, marine organism that produce calcium carbonate shells or skeletons are negatively affected. Coral reefs and coralline algae abilities to produce skeleton also reduces.

Calcium carbonate dissolves in acid. Thus, the more acidic the ocean water is the faster and easier it is to dissolve the exoskeleton and shell of marine organisms made up of calcium carbonate

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3 years ago

PLS ASAP!! Which has a wavelength of 350 nanometers?

slava [35]

Answer:

Ultraviolet light.

Hope this helps!

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3 years ago

Read 2 more answers

A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch

kozerog [31]

Answer:

$\large \boxed{\text{4.63 atm}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

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

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ = 1 °C

p₂ = ?; V₂ = 416 mL; n₂ = n₁; T₂ = 82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}$

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2 years ago

You are trying to determine the volume of the balloon needed to match the density of the air in the lab. You know that if you ca

9966 [12]

To solve this problem, we must be given first the density of air at 20 degrees Celsius. Looking up online, this is equal to:

density air (20C) = 0.0012041 g/mL

so that the volume is:

volume balloon = 0.57 g / (0.0012041 g/mL)

<span>volume balloon = 473.38 mL</span>

5 0

3 years ago