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3 years ago

Why did sir isaac newton refuse a glass of wine before sitting down to formulate the rules for differentiation?

Chemistry

1 answer:

solong [7]3 years ago

6 0

<h2>Answer:</h2>

"He knew better than to drink and derive"

<h3>Explanation:</h3>

Isaac Newton was known to be a passionately focused introvert who could go without sleep for days and live on bread and wine but refused a glass of wine before formulating the rules for differentiation because he did not want to cloud his thinking.

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Alanna spots a bird in her back yard. The bird is sitting on a tree. Explain how the outer coverings of the bird and tree are di

AfilCa [17]

Answer:

A bird's feathers are supposed to keep the bird warm and help them fly. A tree's bark is supposed to protect the inside of the tree.

Explanation:

If a tree has no bark, the inside of it is exposed. If a bird has no feathers, the bird is cold and cant fly.

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3 years ago

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Which observation shows that a substance is malleable

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Answer:

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3 years ago

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What is the molar concentration of 29 g of Mg(OH)2 dissolved in 1.00 L of water

LekaFEV [45]

Answer: The molar concentration of 29 g of $Mg(OH)_{2}$ dissolved in 1.00 L of water is 0.497 M.

Explanation:

Given: Mass = 29 g

Volume = 1.00 L

Moles is the mass of substance divided by its molar mass. So, moles of $Mg(OH)_{2}$ is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{29 g}{58.32 g/mol}\\= 0.497 mol$

Molarity is the number of moles of a substance divided by volume in liter.

Hence, molarity of the given solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.497 mol}{1.00 L}\\= 0.497 M$

Thus, we can conclude that the molar concentration of 29 g of $Mg(OH)_{2}$ dissolved in 1.00 L of water is 0.497 M.

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3 years ago

Be sure to answer all parts. Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everythi

zimovet [89]

Answer:

a) 79.66 seconds is the half-life of the reaction.

b) It will take $4.776\times 10^1$ seconds for 34% of a sample of an acetone to decompose.

c) It will take $2.537\times 10^2$ seconds for 89% of a sample of an acetone to decompose.

Explanation:

The decomposition of acetone follows first order kinetics

The rate constant of the reaction = k = $8.7\times 10^{-3} s^{-1}$

Half life of the reaction = $t_{1/2}$

For the first order kinetic half life is related to k by :

$t_{1/2}=\frac{0.693}{k}$

$t_{1/2}=\frac{0.693}{8.7\times 10^{-3} s^{-1}}=79.66 s=7.966\times 10^1 s$

79.66 seconds is the half-life of the reaction.

Let the initial concentration of acetone be = $[A_o]$

Final concentration of acetone left after t time = [A]

$A=(100\%-34\%)[A_o]=66\%[A_o]=0.66[A_o]$

For the first order kinetic :

$[A]=[A_o]\times e^{-kt}$

$0.66[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}$

Solving for t;

$t=47.76 s$

It will take 47.76 seconds for 34% of a sample of an acetone to decompose.

Let the initial concentration of acetone be = $[A_o]$

Final concentration of acetone left after t time = [A]

$A=(100\%-89\%)[A-o]=11\%[A_o]=0.11[A_o]$

For the first order kinetic :

$[A]=[A_o]\times e^{-kt}$

$0.11[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}$

Solving for t;

$t=2.537\times 10^2 s$

It will take $2.537\times 10^2$ seconds for 89% of a sample of an acetone to decompose.

3 0

3 years ago

the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?

guapka [62]

Answer : The solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$

Explanation :

The solubility equilibrium reaction will be:

$Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}$

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^{+}]^3[PO_4^{3-}]$

$K_{sp}=(3s)^3\times (s)$

$K_{sp}=27s^4$

Given:

$K_{sp}$ = $2.8\times 10^{-18}$

Now put all the given values in the above expression, we get:

$K_{sp}=27s^4$

$2.8\times 10^{-18}=27s^4$

$s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L$

Therefore, the solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$

3 0

3 years ago