All categories

3 years ago

Which tool would best help, will mark brainlist to first 1

Chemistry

2 answers:

V125BC [204]3 years ago

7 0

C. hand lens would magnify the object.

denis-greek [22]3 years ago

3 0

Answer:

Explanation:

You might be interested in

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

Hno3, a strong acid, is added to shift the ag2co3 equilibrium (equation 7.6) to the right. explain why the shift occurs.

Deffense [45]

Answer 1) When a strong acid like $HNO_{3}$ reacts with $Ag_{2} CO_{3}$ usually the equilibrium shifts to the right because

As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change." So, in the given reaction when $HNO_{3}$ reacts with $Ag_{2} CO_{3}$ it generates carbon dioxide and water as a by product, if we are adding $HNO_{3}$ it will remove some of the $CO_{3}$ molecule from the reaction mixture, which then tends to shift the equilibrium towards right.

Answer 2) The same would be observed in this case, if we replace $HNO_{3}$ with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.

As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing $Ag^{+}$ and $Cl^{-}$ ions from the solution.

7 0

2 years ago

Please select the word from the list that best fits the definitions

Mamont248 [21]

Erm i’m afraid we need the list

7 0

2 years ago

Tron 200g dung dich NaCl 40% voi m gam dung dich NaCl 20% thu duoc dung dich NaCl 25% . tinh m

Sati [7]

Answer:cool

Explanation:radical dude

8 0

2 years ago

How can you tell from the name the types of bonds present in a hydrocarbon?

Darya [45]

Answer:

Alkane

Alkene

Alkyne

Explanation:

Alkane=1 bond (Saturated hydrocarbon)

Alkene= 2 bonds (Unsaturated hydrocarbon)

Alkyne= triple bonds (Unsaturated hydrocarbon)

Formula of Alkane = CnH2n+2

Formula of Alkene = CnH2n

Formula of Alkyne = CnH2n-2

7 0

3 years ago