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3 years ago

I need help with #2. I don't know what law to use for problem.

Chemistry

1 answer:

bagirrra123 [75]3 years ago

4 0

This uses something called the combined gas law. The combined gas law is as follows: (P1*V1/T1) = (P2*V2/T2)
According to question 2, you are given the following values initially:
P1 = 680 mm Hg * (1 atm/760 mm Hg) = 0.895 atm
V1 = 20.0 L
T1 = 293 K
STP or standard temperature and pressure implies that the other values we know are:
P2 = 1 atm
T2 = 273 K
Our unknown is V2
If we plug in our known values into the combined gas law:
(P1*V1/T1) = (P2*V2/T2)
(0.895 atm * 20.0 L)/293K = (1 atm * X liters)/273 K
0.0611 L*atm/K = (1 atm * X liters)/273 K
16.7 L = X liters
Therefore, the volume occupied at STP is 16.7 liters
This makes sense because the gas would occupy a smaller volume at a lower temperature, since the gas would have a lower average kinetic energy.

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Write a balanced chemical equation for the reaction between ammonium

Natasha2012 [34]

Answer:

(NH₄)₃PO₄ (l) + Al(NO₃)₃(l) -----------→ AlPO₄(l) + 3NH₄NO₃(l)

Explanation:

Data Give:

Reaction between ammonium phosphate solution and solution of aluminum nitrate

Write a balanced chemical equation

include phase symbols

Details:

To write a balanced chemical equation we have to know formula units of compounds or molecules

Formula units

ammonium phosphate : (NH₄)₃PO₄

aluminum nitrate: Al(NO₃)₃

ammonium nitrate: NH₄NO₃

Now to write a chemical equation

we have to write the chemical formulas or formula unit of each compound
write the reactant on left side of the arrow
write the product on the right side of the arrow
put a plus sign in 2 reactants and products on each side of the arrow
balance the equation by putting coefficient with compound formula
write the phase symbols on the right corner of the compound formula in brackets

So the Reaction will be

(NH₄)₃PO₄ + Al(NO₃)₃ -----------→ AlPO₄ + NH₄NO₃

Now balance the Chemical equation

(NH₄)₃PO₄ + Al(NO₃)₃ -----------→ AlPO₄ + 3NH₄NO₃

Now write the phase Symbols

(NH₄)₃PO₄ (l) + Al(NO₃)₃(l) -----------→ AlPO₄(l) + 3NH₄NO₃(l)

all compounds in the reaction are in liquid form and soluble in water

*** Note:

There is no aluminum nitrite in chemicals formulas

Also ammonium nitrite can not be used in pure isolated form due to its highly instability

4 0

3 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]

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

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:

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

For $_{17}^{35}\textrm{Cl}$ isotope:

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

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

For $_{17}^{37}\textrm{Cl}$ isotope:

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

I need help with all 4 please help

damaskus [11]

Get the app socratic

4 0

3 years ago

Determine the unknown mineral using the information given below. Be sure to use significant figure rules when doing calculations

Mademuasel [1]

Answer:

Mass: 981.0 g

Density: 5.61 g/cm^3

Hardness: = 2.5 - 3

Unknown material: Chalcocite

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3 years ago

How does energy play a role in different rock types forming? I had this question another time in my test and it's asking the sam

Virty [35]

Answer:

Yes

Explanation:

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3 years ago