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Vlad1618 [11]
3 years ago
11

Enter the net ionic equation, including phases, for the reaction of AgNO3 (aq) and KCl(aq). Refer to the solubility rules as nee

ded.

Chemistry
2 answers:
ollegr [7]3 years ago
5 0

Answer:

Ag+(aq) + Cl-(aq) -> AgCl(s)

Explanation:

Given the next molecular equation:

AgNO3 (aq) + KCl (aq) -> AgCl(s) + KNO3 (aq)

To get the net ionic equation, first write separate ions for soluble electrolytes, those indicated with a '(aq)':

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) -> AgCl(s) + K+(aq) + NO3-(aq)

And then cancel any ion on both sides of the previous equation, as follows:

Ag+(aq) + Cl-(aq) -> AgCl(s)

artcher [175]3 years ago
3 0

Answer:

Explanation: you need to remove the subscript of 5

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The properties of a substance are not affected by chemical reactions.<br> O True<br> O False
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Answer:

False

Explanation:

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2 years ago
A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga
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Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0
3 years ago
What is a level 7-8 grade science fair project I can do?
Delicious77 [7]
Making an atom or a virus structure
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3 years ago
A buffer solution is composed of 7.50 mol of acid and 4.75 mol of the conjugate base. If the pKa of the acid is 4.90 , what is t
Mamont248 [21]

Answer:

The correct answer is: pH= 4.70

Explanation:

We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:

pH= pKa + log   \frac{ [conjugate base]}{[acid]}

Given:

pKa= 4.90

[conjugate base]= 4.75 mol

[acid]= 7.50 mol

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7 0
3 years ago
A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere
kvv77 [185]
Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m =  the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
I_{g} =  \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} =  \frac{m}{12} \, kg-m^{2}

According to the parallel axis theorem, the moment of inertia about the first axis is
I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} =  \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
\frac{I_{g}}{I_{1}} =  \frac{0.0833m}{0.1233m} =0.6756

Answer:  0.676

6 0
3 years ago
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