What is the wavelength of a photon containing 3.87x10-23) of energy? (answer in standard form)

How do you know sodium is highly reactive?

Stells [14]

<span>Na (sodium) is highly electropositive. Its has 1 electron in its outermost orbit which is transferred to an electronegative atom to form an ionic bond.

It only needs to get rid of one valence electron to take part in a reaction. That's how it's highly reactive.</span>

4 0

4 years ago

Use complete sentences to differentiate between acids and bases on the basis of touch. Give an example of each type.

stellarik [79]

<h2>Answer:</h2>

Acids are sour, gives burning sensation, generally sticky, reacts with metals to produce hydrogen gas.

example: Acetic acid

Bases are opposite as they are bitter, generally odorless (except ammonia), they are slippery;

example: sodium bicarbonate

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6 0

3 years ago

What volume of NaOH(aq) would be needed to reach the equivalence point if the concentration of the added base were 0.290 M?

DiKsa [7]

Answer:

17.2mL are needed

Explanation:

<em>The pH curve for titration of 50.0 mL of a 0.100 M solution of hydrochloric acid</em>

<em />

A strong acid as HCl reacts with a strong base as NaOH producing water and a salt. The chemical equation is:

HCl + NaOH → H₂O + NaCl

<em>Where 1 mole of HCl reacts per mole of NaOH</em>

<em />

To solve this question we must find the moles of HCl added = Moles NaOH you must add to reach the equivalence point. With these moles and the molar concentration we can find the volume that would be needed as follows:

<em>Moles HCl:</em>

50.0mL = 0.0500L * (0.100moles / L) = 0.00500 moles HCl = Moles NaOH

<em>Volume NaOH:</em>

0.00500 moles NaOH * (1L / 0.290moles) = 0.0172L NaOH =

<h3>17.2mL are needed</h3>

3 0

3 years ago

The mass of an object is 9.505 g and

Goryan [66]

Answer:

The answer is 3 significant figures

7 0

3 years ago

A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.

jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $K_2CO_3$ :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of $K_2CO_3$ :

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles of $K_2CO_3$ = 0.004 moles

For $Ba(NO_3)_2$ :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of $Ba(NO_3)_2$ :

$Moles=0.400\times {30.0\times 10^{-3}}\ moles$

Moles of $Ba(NO_3)_2$ = 0.012 moles

According to the given reaction:

$K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}$

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate = 0.004 moles

Available moles of barium nitrate = 0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0

3 years ago