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3 years ago

What conflicts did issac newton have with Robert Hooke?

Chemistry

1 answer:

Minchanka [31]3 years ago

3 0

Answer:

The great confrontation between the two men occurred in 1686 when Newton published the first volume of his Principia and Hooke affirmed that it was he who had given him the notion that led him to the law of universal gravitation. Hooke demanded credit as the author of the idea but Newton denied it

Explanation:

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The half-life of 131I is 8.021 days.What fraction of a sample of 131I remains after 24.063 days

Leno4ka [110]

24.063/8.021 = 3 half lives
(1/2)^3 =1/8

5 0

3 years ago

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or io

Dmitrij [34]

Answer :

(a) The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

The spectator ions are, $NH_4^{+}\text{ and }SO_4^{2-}$

(b) The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

(c) The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

The spectator ions are, $K^{+}\text{ and }NO_3^{-}$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The given balanced ionic equation will be,

$Cr_2(SO_4)_3(aq)+3(NH_4)_2CO_3(aq)\rightarrow 3(NH_4)_2SO_4(aq)+Cr_2(CO_3)_3(s)$

The ionic equation in separated aqueous solution will be,

$2Cr^{2+}(aq)+3SO_4^{2-}(aq)+6NH_4^{+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)+6NH_4^{+}(aq)+3SO_4^{2-}(aq)$

In this equation, $NH_4^{+}\text{ and }SO_4^{2-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2Cr^{2+}(aq)+3CO_3^{2-}(aq)\rightarrow Cr_2(CO_3)_3(s)$

(b) The given balanced ionic equation will be,

$Ba(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow 2KNO_3(aq)+BaSO_4(s)$

The ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

(c) The given balanced ionic equation will be,

$Fe(NO_3)_2(aq)+2KOH(aq)\rightarrow 2KNO_3(aq)+Fe(OH)_2(s)$

The ionic equation in separated aqueous solution will be,

$Fe^{2+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)$

In this equation, $K^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Fe^{2+}(aq)+2OH^{-}(aq)\rightarrow Fe(OH)_2(s)$

3 0

3 years ago

Calculate (a) the number of moles of hydrogen required to react with 0.0969 moles of nitrogen, and (b) the number of moles of am

Bumek [7]

Answer:

The answer to your question is:

Explanation:

Data

moles H=?

moles of N = 0.0969

moles of NH₃=?

N₂ (g) + 3 H₂ (g) ⇒ 2NH₃ (g)

Process

1.- Set a rule of three to calculate the moles of hydrogen

1 mol of nitrogen ------------- 3 moles of hydrogen

0.0969 moles of N ---------- x

x = (0.0969 x 3) / 1

x = 0.2907 moles of hydrogen

2.- Set a rule of three to calculate the moles of ammonia

1 mol of nitrogen -------------- 2 moles of ammonia

0.0969 mol of N -------------- x

x = (0.0969 x 2) / 1

x = 0.1938 moles of ammonia

5 0

3 years ago

Which of the following statements about connecting paragraphs is correct?

Reika [66]

B.

And maybe put your question in the English/Literature tag next time lol

4 0

2 years ago

Read 2 more answers

If I use the dropper to add 10 mL of drink mix to 50 mL of water, what is the

Yuliya22 [10]

Answer:

16.7%

Explanation:

From the question given above, the following data were obtained:

Volume of drink mix = 10 mL

Volume of water = 50 mL

Percentage of drink mix =?

Next, we shall determine the total volume of the solution. This can be obtained as follow:

Volume of drink mix = 10 mL

Volume of water = 50 mL

Total volume of solution =?

Total volume of solution = (Volume of drink mix) + (Volume of water)

Total volume of solution = 10 + 50

Total volume of solution = 60 mL

Finally, we shall determine the percentage of the drink mix in the solution. This can be obtained as follow:

Volume of drink mix = 10 mL

Total volume of solution = 60 mL

Percentage of drink mix =?

Percentage of drink mix = Vol. of drink mix / total volume × 100

Percentage of drink mix = 10/60 × 100

Percentage of drink mix = 16.7%

4 0

3 years ago