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4 years ago

A plant dying after being exposed to poison represents a physical change true or false?

Chemistry

2 answers:

Keith_Richards [23]4 years ago

6 0

False actually.

I'm sorry i don't know how i just got it right on my test and i hope you will?did to. And i hope i also get marked brainliest :)

r-ruslan [8.4K]4 years ago

3 0

Yes, i believe that it can have a physical change but it is most likely chemical.

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DUE IN 30 MINUTES!!! (please keep it short if you can, if not its ok.) What is the molar mass of Pb(SO4)2? Explain how you calcu

kipiarov [429]

Answer:

399

Explanation:

Pb(SO4)2 contains 1 atom of Pb, 2 atoms of S and 8 atoms of O. So, atomic mass of Pb(SO4)2 is 207 + 64 + 128 = 399 u. Therefore, molar mass of Pb(SO4)2 is 399 g/mol.HOPE THIS helps. Good Luck

5 0

3 years ago

Read 2 more answers

If I add 2.75 g of CaO to 155 mL of distilled water, what will be the concentration of the solution be?

IRINA_888 [86]

M = m/L
moles of CaO = 2.75g / molar mass
2.75g/ 56.08g = 0.049moles CaO
155mL / 1000 = 0.155 L
M= 0.049moles/0.155L
M=0.316

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2 years ago

How to do balance chemical equiations

alex41 [277]

Hey, lovely! It's a pretty lengthy process but here is a pretty clear video on how to do it. Hope this helps ya!

https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equat...

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4 years ago

Suppose you put equal amounts of pure water and salt water into separate ice cube trays of the same size and shape. What would e

gavmur [86]

Both the sat water ice cubes would freeze due to the temperature of the freezer.

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3 years ago

Read 2 more answers

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago