For which one of the following molecules is the indicated type of hybridization not appropriate for

Lynna [10]

Answer:

$\boxed {\boxed {\sf D. 0.440 \ mol \ NaOH}}$

Explanation:

To convert from moles to grams, the molar mass is used (mass of 1 mole). The values are the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units.

<h3>1. Molar Mass</h3>

We are given the compound sodium hydroxide (NaOH) and we need to look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

The formula for the compound has no subscripts, so there is 1 mole of each element in 1 mole of the substance. We can simply add the molar masses.

NaOH: 22.9897693 + 15.999 + 1.008 = 39.9967693 g/mol

This means there are 39.9967693 grams of sodium hydroxide in 1 mole.

<h3>2. Convert Grams to Moles </h3>

Use the molar mass we found as a ratio.

$\frac {39.9967693 \ g \ NaOH}{ 1 \ mol \ NaOH}$

Since we are converting 17.6 grams of NaOH to moles, we multiply by this value.

$17.6 \ g\ NaOH *\frac {39.9967693 \ g \ NaOH}{ 1 \ mol \ NaOH}$

Flip the ratio so the units of grams of NaOH cancel.

$17.6 \ g\ NaOH *\frac {1 \ mol \ NaOH}{ 39.9967693 \ g \ NaOH}$

$17.6 *\frac {1 \ mol \ NaOH}{ 39.9967693}$

$\frac{17.6 }{ 39.9967693} \ mol \ NaOH$

$0.4400355406 \ mol \ NaOH$

<h3>3. Round </h3>

The original measurement of grams has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

0.4400355406

The 0 in the ten thousandths place (in bold above) tells us to leave the 0 in the thousandth place.

$0.440 \ mol \ NaOH$

17.6 grams of sodium hydroxide are equal to <u>0.440 moles of sodium hydroxide.</u>

3 0

3 years ago

To separate a mixture of iron filings and salt, the most efficient method would be

GaryK [48]

Answer:

Explanation:

Of course you could do the separation chemically. Dissolve the salt up in water, pass thru a filter, wash the iron filings with ethanol, which would encourage the salt to precipitate from solution.

I do hope I helped you! :)

4 0

3 years ago

What is the representative unit in a molecular compound

Anastasy [175]

Answer:

A.) A molecule

Explanation:

3 0

2 years ago

Read 2 more answers

Given pH = 3.50 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?

lys-0071 [83]

Answer:

Explanation:

Given parameters:

pH = 3.50

Unknown:

concentration of [H₃0⁺] = ?

concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

pH = -log₁₀[H₃O⁺]

[H₃O⁺] = inverse log₁₀ (-pH) = $10^{-pH}$ = $10^{-3.5}$

[H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

For the [OH⁻]:

we use : pOH = -log₁₀ [OH⁻]

Recall: pOH + pH = 14

pOH = 14 - pH = 14 - 3.5 = 10.5

Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

[OH⁻] = $10^{-pOH}$

[OH⁻] = $10^{-10.5}$ = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

8 0

3 years ago

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

3 years ago