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2 years ago

Which statement is correct for the element of proton number 19?

Chemistry

1 answer:

BlackZzzverrR [31]2 years ago

8 0

Answer:

Explanation:

From periodic table : proton number 19 corresponds to K (potassium)

which is very reactive with water

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2 SO3 (g) + Heat <-----> 2 SO2 (g) + O2 (g)

Step2247 [10]

Answer:

The concentration of SO₂ will decreases

Explanation:

As you can see in the reaction

2 moles of gas ⇆ 3 moles of gas

Based on Le Châtelier's principle, a change doing in a system will produce that the system reacts in order to counteract the change made.

If the pressure is increased, the system will shift to the left in order to produce less moles of gas and decrease, thus, the pressure.

As the system shift to the left, the concentration of SO₂ will decreases

7 0

3 years ago

What type of energy does water gain as it sits in the sun?

Sveta_85 [38]

Answer:

A. Heat energy

Explanation:

7 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago

ANSWER QUICK PLEASE WORTH 100 POINTS

sukhopar [10]

Answer:

Explanation:

Volume is defined as the space occupied by an object or substance irrespective of its state of matter.The conversion used from millimeter to liter is:

1 milliiliter = 0.001 L

Therefore, we can convert the volume of sample from 2.5 ml in liters as follows.

2.5 ml in liters = 2.5ml x 0.001 L/1ml

= 0.0025 L

Thus, we can conclude that the volume of given sample in liter is 0.0025 L

Hope this helps! :)

6 0

2 years ago

Plz answer I need the answer What is the goal of technology?

Sloan [31]

I think the answer is C

7 0

3 years ago

Read 2 more answers