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3 years ago

How do decantation and filtration differ which should be faster

1 answer:

8 0

Decantation describes when a substance settles to the bottom, think of the Hershey syrup and milk, the syrup slowly settles to the bottom. Filtration describes a mixture and as an example Sand and Water and separation comes from setting up a filter paper. Filtration is usually faster.

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How many grams are there in 3.30x10 23 grams of n2i6

qwelly [4]

Technically, the answer should be 3.30 * 10^23 grams. But I think you mean either molecules, atoms, moles or grams of Ni2I6 with that number of molecules .

1 mole of Ni2 I6 = 6.02 * 10^23 molecules
x [mole] = 3.30 * 10^23 molecules

1/x = 6.02 * 10^23 / 3.30 * 10^23 Cancel the 10^23 on the right side
1/x = 6.02 / 3.30 Cross multiply
3.30 = 6.02 x Divide by 6.02
3.30 / 6.02 = x
x = 0.548 moles

what to do from here?

1 mole of Ni2I6 is
2 * Ni = 2 * 59 = 118 grams
6 * I = 6 * 131 = 786 grams
Total = 904 grams

Set up a proportion.
1 mole Ni2I6 = 904 grams
0.548 moles = x

1/0.548 = 904/x Cross multiply
x = 0.548 * 904
x = 495.4 grams of Ni2I6 <<<<<< Answer.

8 0

3 years ago

What is the most metallic character with low ionization energies

Nat2105 [25]

Answer:

Metals tend to lose electrons in chemical reactions, as indicated by their low ionization energies. Within a compound, metal atoms have relatively low attraction for electrons, as indicated by their low electronegativities.

4 0

2 years ago

Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O

Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = $\frac{15.0 g}{40 g/mol}=0.375 mol$

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

$Molarity=\frac{Moles}{Volume(L)}$

$n=0.250 M\times 0.150 L=0.0375 mol$

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

$\frac{1}{1}\times 0.0375 mol$ of NaOH

Moles of NaOH left unreacted in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

$1\times 0.3375 moles=0.3375 mol$ of hydroxide ion

The molarity of hydroxide ion in solution ;

$=\frac{0.3375 mol}{0.150 L}=2.25 M$

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0

3 years ago

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

A scientist obtains a set of measurements that are very close to one another. These can be said to be:

pickupchik [31]

Answer:

I believe this is a K-12 test question. If the answers below are what you have on your test . . .

- Precise

- Accurate

- Identical

- None of the above

Then the answer is precise.

3 0

2 years ago