All categories

3 years ago

What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 11.630 g to 4.000 g ?

Chemistry

1 answer:

Dima020 [189]3 years ago

5 0

The amount of sugar is 2621 mg

Why?

The complete question is:

A 12.630 g milk chocolate bar is found to contain 8.315 g of sugar.

Part B. What will be the amount of sugar in milligrams if the size of the milk chocolate bar is reduced from 12.630 g to 4.000 g ?

To find the answer we have to determine first the amount of sugar in milligrams per gram of chocolate bar. We can find that by applying the following conversion factor:

$\frac{8.315 gSugar}{12.630 g Chocolate}*\frac{1000mg}{1g}=658.35mgSugar/gChocolate$

Now, we have to determine the amount of sugar in milligrams if we had a chocolate bar with 4.000 g:

$4.000gChocolate*\frac{658.35mgSugar}{1gChocolate}=2621mgSugar$

Have a nice day!

You might be interested in

A chemist wants to combine sodium with a halogen. The halogen must have a higher atomic mass than sodium. Which of the following

user100 [1]

Answer:

bromine

Explanation:

i just know

8 0

3 years ago

Deuterium is a relatively uncommon form of hydrogen, but can be created from what common source?.

Elena-2011 [213]

Deuterium is a relatively uncommon form of hydrogen, but can be created from water.

Heavy hydrogen commonly known as deuterium
stable isotopes of hydrogen
gets its name from the Greek word deuterons means second.
has only one proton and one neutron
nucleus of the hydrogen's deuterium atom is known as a deuteron containing one proton and one neutron.
Deuterium forms chemical bonds that are stronger than regular hydrogen
gas deuterium is colorless
Deuterated water is used in Magnetic Resonance Spectroscopy.
used in the determination of the isotopologue of various organic compounds.
used in Infrared Spectroscopy.

To know more about Deuterium visit : brainly.com/question/27870183

#SPJ4

8 0

1 year ago

When the u-235 nucleus is struck with a neutron, the ce-144 and sr-90 nuclei are produced, along with some neutrons and beta par

butalik [34]

Answer: 2 (2 neutrons are produced).

Explanation:

1) In the left side of the transmutation equationa appears:
²³⁵U + ¹n →

I am omitting the atomic number (subscript to the leff) because the question does not show them as it is focused on number of neutrons.

2) The right side of the transmutation equation has:

→ ¹⁴⁴Ce + ⁹⁰Sr + ?

3) The total mass number of the left side is 235 + 1 = 236

4) The total mass number of Ce and Sr on the right side is 144 + 90 = 234

5) Then, you are lacking 236 - 234 = 2 unit masses on the right side which are the 2 neutrons that are produced along with the Ce and Sr.

The complete final equation is:

²³⁵U + ¹n → ¹⁴⁴Ce + ⁹⁰Sr + 2 ¹n

Where you have the two neutrons produced.

6 0

3 years ago

Iron 3 oxide and carbon react to form iron and carbon dioxide. Balance the equation.

Oksi-84 [34.3K]

Answer:

2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2

Explanation:

First of all, you have to translate the words into an equation.

Fe(iii)2O3 + C ==> Fe + CO2

The easiest way to tackle this is to start with the Oxygens and balance them. They must balance by going to the greatest common factor which is 6. So you multiply the molecule by whatever it takes to get the Oxygens to 6

2 Fe(iii)2O3 + C ==> Fe + 3 CO2

Now work on the irons. There 2 on the left and just 1 on the right. So you need to multiply the iron by 2.

2 Fe(iii)2O3 + C ==> 2 Fe + 3 CO2

Finally it is the turn of the carbons. There are 3 on the right, so you must make the carbon on the left = 3

2 Fe(iii)2O3 + 3 C ==> 2 Fe + 3 CO2

And you are done.

5 0

3 years ago

This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by <em>'n'</em>.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0

2 years ago

Read 2 more answers