Ask question

Ask question

All categories

3 years ago

10

A baker places a paper cone into the center of a fruit pie before putting the pie in the oven. Explain why this would keep juice

from running out of the pie during baking

1 answer:

jonny [76]3 years ago

5 0

Because the paper absorbs the juice?

You might be interested in

If more solute can be dissolved in a solvent, the solution is

WARRIOR [948]

Answer:

the answer is unsaturated

Explanation:

A saturated solution contains more solute per volume of solvent than an unsaturated solution. The solute has dissolved until no more can, leaving undissolved matter in the solution. ... In a supersaturated solution, there is more dissolved solute than in a saturated solution.

7 0

3 years ago

QUICKEST GETS BRAINLIEST AND 7 POINTS

Angelina_Jolie [31]

Hello there.

Thomson's atomic model is best described by which of the following statements?

A nucleus with electrons moving around it like planets.

3 0

3 years ago

Anyone know how to hide a body asking for a friend

Wittaler [7]

Throw it in the ocean lol

4 0

2 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

<u>Answer:</u> The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

<u>Explanation:</u>

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:

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

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

<u>For $_{17}^{35}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

<u>For $_{17}^{37}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago

Please someone help I’m really confused

NARA [144]

Answer:10-3

Explanation:

7 0

2 years ago