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3 years ago

11

Food couldn’t reach the stomach without the esophagus and the _? please help only correct answers.

1 answer:

expeople1 [14]3 years ago

3 0

<span>Food couldn’t reach the stomach without the esophagus and the throat.</span>

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URGENT! Consider the following decomposition reaction of ammonium carbonate ((NH4)2CO3):

dexar [7]

<span>Using PV=nRT to find the moles and then convert back.
</span><span>4x=.8944
</span><span>solve for x then use the pressure for lets say CO2 put that into PV=nRT then solve for n then convert over.
</span>
<span>(.2236)(2)/(298*.08206) = .0183*96g/mol = 1.76g
</span>
<span>For C:

[NH3]^2[CO2][H2O] = Kp
x=0.2236 (2*.2236)^2(.2236)*(.2236)
=0.001
</span>

6 0

2 years ago

A student puts a glass of water in the freezer. Later, he notices ice forming on the surface of the water. Which property of wat

Karolina [17]

C.ice has a lower freezing point than liquid water

hope this helps

4 0

2 years ago

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Scientists claim to be the way the continents were arranged 250 million years ago. Scientists who make this claim call this anci

Mkey [24]

Answer:

I dn't mnoefbv

Explanation:

Gr

6 0

2 years ago

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In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2

allsm [11]

Answer:

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

Explanation:

$SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)$

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance the charge by adding 2 electrons on product side:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-$ ....[1]

Reduction :

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)$

Balance O by adding water on required side:

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

Now, balance H by adding $H^+$ on the required side:

$SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

At last balance the charge by adding electrons on the side where positive charge is more:

$SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)$ ..[2]

Adding [1] and [2]:

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

4 0

2 years ago

Convert 125m^3 to km^3

Ugo [173]

Answer:

1.25e-7

Explanation:

4 0

2 years ago