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3 years ago

The j shaped organ is full of hydrochloric acid

1 answer:

8 0

This would be a stomach.

The stomach is a muscular, J-shaped organ in the upper part of the abdomen. It is part of the digestive system, and serves to break down food through dissolving by acid.

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Imagine the reaction A + B ⇌ C + D is at equilibrium, where the forward and reverse reactions are at equal rates. What would hap

kondaur [170]

Answer:

Rate of forward reaction will increase.

Explanation:

Effect of change in reaction condition on equilibrium is explained by Le Chatelier's principle. According to this principle,

If an equilibrium condition of a dynamic reversible reaction is disturbed by changing concentration, temperature, pressure, volume, etc, then reaction will move will in a direction which counteract the change.

In the given reaction,

A + B ⇌ C + D

If concentration of A is increase, then reaction will move in a direction which decreases the concentration of A to reestablish the equilibrium.

As concentration A decreases in forward direction, therefore, rate of forward reaction will increase.

3 0

3 years ago

Can somebody answer this for me

valkas [14]

Answer: C (Option 3)

Cs < Cl < F

Cesium is the least electronegative atom.

Fluorine is the most electronegative atom.

5 0

3 years ago

4) What volume will the gas in the balloon at right occupy at 250k? balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$