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stiv31 [10]
3 years ago
5

Write the net ionic equation, including the phases.2hno3(aq)+sr(oh)2(aq) ---> 2h2o(l)+srno32(aq)

Chemistry
1 answer:
Nezavi [6.7K]3 years ago
7 0
Balanced chemical reaction: 
2HNO₃<span>(aq) + Sr(OH)</span>₂(aq) → 2H₂O(l) + Sr(NO₃)₂<span>(aq).
Balanced ionic reaction:
2H</span>⁺(aq) + 2NO₃⁻(aq) + Sr²⁺(aq) + 2OH⁻(aq) → 2H₂O(l) + Sr²⁺(aq) + 2NO₃⁻(aq).
Net ionic reaction:
2H⁺(aq) + 2OH⁻(aq) → 2H₂O(l).
(aq) means that substances are dissociated on cations and anions in water.
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A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under
sukhopar [10]

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

8 0
3 years ago
Watch the animation depicting Rutherford’s experiment and choose which of the following conclusions are correct.a. The atom cont
Lilit [14]

Answer:

a, b, c, d

Explanation:

Rutherford’ atomic model is based on the gold foil experiment. In this experiment, beam of alpha rays was bombarded on thin gold foil. He observed that:

Most of the alpha particles passed through thin foil without any deflection.

Few alpha particles deflected by an angle of 90o.

Based on observation, Rutherford concluded that majority of the space inside the atom is empty.

He explained defection of few alpha particles by assuming that most of the mass is concentrated at the nucleus and positively charged.  

Therefore, among given, the correct statements are:

The atom contains a positively charged nucleus.

Positive charge is condensed in one location within the atom.

The majority of the space inside the atom is empty space

The mass of an atom is concentrated at the nucleus

Therefore, the correct options are:

a, b, c, d

4 0
3 years ago
Create the Lewis structure of CF4. NOTE: do only the step asked for in each part and then click Check–don't work ahead to solve
Naya [18.7K]

To create the Lewis structure we need to take into account the octet rule: atoms tend to gain, lose or share electrons to complete their valence shell with 8 electrons.

C belongs to Group 4A in the periodic table so it has 4 valence electrons. It needs to share 4 pairs of electrons to complete the octet.

F belongs to Group 7A in the periodic table so it has 7 valence electrons. Each F needs to share 1 pair of electrons to complete the octet.

As a consequence, in CF₄, C will form a single bond with each F and all the octets will be complete.

5 0
3 years ago
STATION 1
grigory [225]

Answer:

B - (C , Al, P, Cl)

Explanation:

How I got this answer was by looking at my periodic table it shows you how much it contains by the Atomic number.

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Atomic number on P (Phosphorus) is - 15

Atomic number on Cl (Chlorine) is - 17

Now it says least to greatest and the other options are wrong I did the work for you hope this helps :)) I also had this project  you didnt ask but the answer for the The Lesson are {B E M S} which as the code numbers are gonna be -7494- Im glad to help if you need more help I will give you the other answers as well :) !

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3 years ago
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victus00 [196]

<h2>Answer with explanation </h2>

<h3><em>The starting diol for this molecule is :-</em></h3><h3><em>The starting diol for this molecule is :-D) ethan-1,2-diol.</em></h3>

<em>Hope </em><em>my </em><em>answer </em><em>is</em><em> helpful</em><em> to</em><em> you</em><em> </em><em>☺️</em>

3 0
3 years ago
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