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Luda [366]
2 years ago
11

How much time is needed to form most fossils? Help please!

Chemistry
2 answers:
Varvara68 [4.7K]2 years ago
5 0

Answer:millions of years

Explanation:

Edge 2021

Hunter-Best [27]2 years ago
3 0

Answer:

a minimum of <em>1</em><em>0</em><em>,</em><em>0</em><em>0</em><em>0</em><em> </em>years

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1 point
inysia [295]

Answer:

The fungus has grown larger

Explanation:

Because where the orange is in the fridge and even normally you out oranges on the counter or in a bowl, where it's in the fridge it got old faster.

8 0
3 years ago
What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?
tensa zangetsu [6.8K]

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

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3 0
3 years ago
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Paul [167]

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6 0
1 year ago
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NaF + HCL=&gt; NACL + HF
Jobisdone [24]

Answer:

NaF + HCl --> NaCl + HF

Explanation:

hope it helps you

4 0
2 years ago
in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc
Margarita [4]

Answer:  0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}     .....(1)

a) Molarity of Na_2S solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol

\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol

b) Molarity of Hg(NO_3)_2 solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol

Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3

According to stoichiometry :

1 mole of Hg(NO_3)_2 reacts with 1 mole of Na_2S

Thus 0.0005 moles of HgNO_3 reacts with=\frac{1}{1}\times 0.0005=0.0005 moles of Hg(NO_3)_2

Thus Hg(NO_3)_2 is the limiting reagent and Na_2S is the excess reagent.

According to stoichiometry :

1 mole of Hg(NO_3)_2 forms=  1 mole of Hg_2S

Thus 0.0005 moles of Hg(NO_3)_2 forms=\frac{1}{1}\times 0.0005=0.0005 moles of Hg_2S

mass of H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0
3 years ago
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