Please help with one and 2

A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man

kap26 [50]

<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

Mass of NaBr sample is 11.97 g

% composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

A sample of 11.97 g of NaBr contains 22.34% of Na by mass

Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

= 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

3 0

3 years ago

What type of bond is made between amino acids in the ribosome?

sleet_krkn [62]

Answer:

<em><u>Peptide bonds form between the amino group of the amino acid attached to the A-site tRNA and the carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide bond is catalyzed by peptidyl transferase, an RNA-based enzyme that is integrated into the 50S ribosomal subunit.</u></em>

8 0

3 years ago

Which solution would most likely cause a plant placed in it to become firmer and more rigid? A. hypertonic B. hypotonic C. isoto

Anna11 [10]

"Hypotonic" is the one solution among the choices given in the question that would <span>most likely cause a plant placed in it to become firmer and more rigid. The correct option among all the options that are given in the question is the second option or option "B". I hope the answer has come to your great help.</span>

5 0

3 years ago

Read 2 more answers

What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch

sammy [17]

Explanation:

It is known that charge on xenon nucleus is $q_{1}$ equal to +54e. And, charge on the proton is $q_{2}$ equal to +e. So, radius of the nucleus is as follows.

r = $\frac{6.0}{2}$

= 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

d = r + 2.5

= (3.0 + 2.5) fm

= 5.5 fm

= $5.5 \times 10^{-15} m$ (as 1 fm = $10^{-15}$ )

Therefore, electrostatic repulsive force on proton is calculated as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}$

Putting the given values into the above formula as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}$

= $(9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}$

= $(9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}$

= 411.2 N

or, = $4.1 \times 10^{2}$ N

Thus, we ca conclude that $4.1 \times 10^{2}$ N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0

4 years ago

An ice cube and a scoop of table salt are left outside on a warm sunny day. explain why the ice cube melts and the salt does not

Anna11 [10]

The melting point of ice is 0 degrees Celcius, which means it exists as a liquid for any temperatures above 0 degrees. The melting point of salt is approximately 800 degrees Celcius, which is way greater than the melting point of ice. This means that for temperatures below 800 degrees, salt exists as a solid.

The temperature of the area where they were placed we can assume was somewhere between 0 and 800 degrees, greater than the melting point of ice but less than the melting point of salt. This why the ice melted but the salt did not.

I hope this helps!

3 0

3 years ago