Organic compounds are distinguished by molecules that contain _______

Ostrovityanka [42]

The answer is B. It will grow towards the hole in the box because the plant will try and get as close to the sunlight as possible which is called phototropism.

5 0

4 years ago

Read 2 more answers

Give six examples of complex compounds.

ra1l [238]

Answer:

Examples of complex compound include potassium ferrocyanide K4[Fe(CN)6] and potassium ferricyanide K3[Fe(CN)6]. Other examples include pentaamine chloro cobalt(III) chloride [Co(NH)5Cl]Cl2 and dichlorobis platinum(IV) nitrate [Pt(en)2Cl2](NO3)2.

3 0

4 years ago

How many grams of 48.0 wt% NaOH (FM 40.00) should be diluted to 1.00 L to make 0.11 M NaOH? (Enter your answer using two digits.

kifflom [539]

Answer:

The answer is 916.67 g

Explanation:

48.0 wt% NaOH means that there are 48 g of NaOH in 100 g of solution. With this information and the molecular weight of NaOH (40 g/mol), we can calculate the number of mol there are in 100 g of this solution:

$\frac{48 g NaOH}{100 g solution}$ x $\frac{1 mol NaOH}{40 g}$ = 0.012 mol NaOH/100 g solution

Finally, we need 0.11 mol in 1 liter of solution to obtain a 0.11 M NaOH solution.

0.012 mol NaOH ------------ 100 g solution

0.11 mol NaOH------------------------- X

X= 0.11 mol NaOH x 100 g/ 0.012 mol NaOH= 916.67 g

We have to weigh 916.67 g of 48.0%wt NaOH and dilute it in a final volume of 1 L of water to obtain a 0.11 M NaOH solution.

7 0

3 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is $1.3\times 10^2g$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

5 0

3 years ago

What type of map projection makes plotting travel by airplane the easiest? I know it's azimuthal maps but why

KatRina [158]

Answer:

Explanation:

Azimuthal maps projection is the easiest for plotting travel by airplane because Azimuthal maps make the UV plane to be tangent to the globe and it's make the directions from the center of projections to be correct and accurate to any other points , likewise circles are also projected directly to the straight lines on the plane. And this is because Any line that is normally drawn through the tangent point give distance correctly and accurately.

7 0

3 years ago

Organic compounds are distinguished by molecules that contain ________ bonded to other elements.