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Kisachek [45]
3 years ago
8

9.44 x 10^3 In standard form

Chemistry
1 answer:
Marta_Voda [28]3 years ago
5 0
9440 is standard form.
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Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -&gt; 1 ______ +2LiF
ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0
3 years ago
Phosphorus has three electrons in its 3p sublevel and sulfur has four. Phosphorus should have the lower ionization energy but it
Nadya [2.5K]

Answer:

B

Explanation:

First of all it is important to know that a half filled orbital is particularly stable. In phosphorus all the electrons occur singly in the 3p sublevel minimizing inter electronic repulsion hence it is more difficult to remove an electron from this energetically stable arrangement. In sulphur, electrons are paired in one of the 3p orbitals thereby lowering the energy of that level due to instability caused by interelectronic repulsion between two electrons in the same orbital.

8 0
2 years ago
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vazorg [7]

^{hello}.

Sorry, I'm am not going to entertain you!

But, have a great day!

#LearnWithBrainly

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TanakaBro

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Calculate the work associated with the compression of a gas from 121.0 L to 80.0 L at a constant pressure of 16.7 atm.
Mkey [24]

Answer:

6.94 × 10⁴ J

Explanation:

We can calculate the work (W) associated with the compression of a gas at constant pressure using the following expression:

W = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

Knowing that 1 atm.L = 101.325 J, the work of compression is:

W=-P\times \Delta V=-16.7atm\times(80.0L-121.0L)\times\frac{101.325J}{1atm.L} =6.94\times10^{4} J

The positive sign of the work means that the surroundings do work on the system.

4 0
2 years ago
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