What are the tracings of seismometers called

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(aq) What

ryzh [129]

Answer : The mass of $PbI_2$ precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of $PbI_2$ = 461.01 g/mole

First we have to calculate the moles of $NaI$ .

$\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles$

Now we have to calculate the moles of $PbI_2$ .

The balanced chemical reaction is,

$Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaI$ react to give 1 mole of $PbI_2$

So, 0.042 moles of $NaI$ react to give $\frac{0.042}{2}=0.021$ moles of $PbI_2$

Now we have to calculate the mass of $PbI_2$ .

$\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2$

$\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g$

Therefore, the mass of $PbI_2$ precipitate produced will be, 9.681 grams.

6 0

4 years ago

An experiment was set up to test the hypothesis that all plants will grow faster in a "super soil" versus regular potting soil.

dolphi86 [110]

Answer:

add more types of plants

Explanation:are you kidding me i’m doing this stuff and i’m in 7th grade

5 0

4 years ago

calculate the molar mass of a gas if 3.30 grams of the has occupies 660 ml at 735mmHg and 27 degrees C

AnnZ [28]

Use the ideal gas law PV = nRT and manipulate it to solve for n. PV/RT = n. Convert to SI units: 660ml -> 0.660 L 27 degrees -> 300 K Leave Pressure as is Use 62.36368 for R becuase you are using mmHg for Pressure. Now plug into the equation to get about 0.025928 mols. Divide the 3.30 grams by the mols to get about 127 g/mol.

8 0

3 years ago

Describe examples of failed experiments that produced valuable information

Nataliya [291]

Hello Norgepr, examples of failed experiments that produced valuable information can be the mage via BoingBoing There have been ample stories of human children being raised by other species and eventually becoming more like that animal than an actual human. If the process could go one way, Winthrop Kellogg was sure that it could also go the other, particularly if the animal involved was one of our closest genetic cousins. In 1931, Kellogg received a grant for his experiment and the timing couldn’t be better –his wife just had a baby boy, David. This would give them the unique opportunity to raise a baby chimp, named Gua, right along side a human baby. It didn’t take long for the babes to bond and become best friends. Kellogg and his wife took impeccable notes on their two “children” noting their physical changes, emotions and how they scored on small intelligence tests. The chimp scored notably higher on the intelligence tests due the fact that the species matures faster than human babies. Gua picked up quite a few human behaviors, such as walking upright and eating with a spoon, but she failed to learn how to speak and learn simple repetition games, like patty cake. Her emotions were also much less predictable and inclined to change at the drop of a hat. Unfortunately, the experiment really started to go wrong when little David started to become more chimplike than Gua became humanlike. He only learned a few simple words and often took to making chimp howls when he wanted something. After only nine months, the Kelloggs gave up on Gua, concerned that David would fail to grow up like a normal human child. In the years since this project, plenty of people have adopted chimps as babies, proving beyond a doubt that the animals can never act completely human –even if they are adorable in overalls.

3 0

3 years ago

A compound sample contains 60.87% c, 4.38% h, and 34.75% o by mass. it has a molar mass of 276.2 g/mol. what are the empirical a

Allushta [10]

Step 1) Assume we have 100 grams of the compound. Thus, we're starting with 60.87 grams C, 4.38 grams H, and 34.75 grams O.

Step 2) Convert the masses of each to moles.

60.87 grams C=5.068276436 mol C

4.38 grams H=4.345238095 mol H

34.75 grams O=2.171875 mol O

step 3) Determine your simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles. (In this case, the smallest number of moles is 2.171875 mol O)

O: $\frac{2.171875}{2.171875} = 1$

H: $\frac{4.345238095}{2.171875} = 2$

C: $\frac{5.068276436}{2.171875} = 2.33$

Step 4) Whenever you're doing an empirical formula problem, and you run into a number that ends with .33, rather than rounding down to a whole number, you need to multiply all your ratios by 3 to obtain whole numbers. Thus, you will get:

O=3

H=6

C=7

Step 5) write the empirical formula using the ratios.

The empirical formula is: C₇H₆O₃

Step 6) The subscripts in the molecular formula of a substance are always whole-number multiples of the subscripts in its empirical formula. This multiple is found by dividing the molecular weight (which was given to us in the problem: 276.2 amu) by the empirical formula weight (C₇H₆O₃ =138.118 amu).

$\frac{276.2}{138.118} = 2$

Step 7) simply multiply the subscripts in the empirical formula by the multiple, 2, and you will get the molecular formula.

The molecular formula is: C₁₄H₁₂O₆

6 0

3 years ago