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AlexFokin [52]
3 years ago
6

Place the events in order as the temperature of a gas in a container slowly increases . I will reward brainliest

Chemistry
2 answers:
Tamiku [17]3 years ago
7 0
2, 3, 1, 5, 4.. i believe

Levart [38]3 years ago
5 0
Ghjjjjn just here for points
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Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)
Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(c) K⁺ (charge 1+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0
3 years ago
A copper rod that has a mass of 200.0 g has an initial temperature of 20.0°C and is heated to 40.0°C. If 1,540 J of heat are nee
almond37 [142]

Answer:

Specific heat of copper = 0.385 J/(gi°C)

Explanation:

Given:

Mass of copper (m) = 200 g =

Initial temperature (T1) = 20°C

Final temperature (T2) = 40°C

Heat needed (Q) = 1,540 J

Find:

Specific heat of copper = ?

Computation:

⇒ Change in temperature (ΔT) = T2 - T1

⇒ Change in temperature (ΔT) = 40°C - 20°C

⇒ Change in temperature (ΔT) = 20°C

⇒ Specific heat of copper = Q / [mΔT]

⇒ Specific heat of copper = 1,540 / [(200)(20)]

Specific heat of copper = 0.385 J/(gi°C)

5 0
3 years ago
Calculate the equilibrium concentration of c2o42− in a 0.20 m solution of oxalic acid.
spin [16.1K]

Answer:

[C2O2−4] =1.5⋅10−4⋅mol⋅dm−3

Explanation:

For the datasheet found at Chemistry Libretext,

Ka1=5.6⋅10−2 and Ka2=1.5⋅10−4 [1]

for the separation of the primary and second nucleon once oxalic corrosive C2H2O4 breaks up in water at 25oC (298⋅K).

Build the RICE table (in moles per l, mol⋅dm−3, or identically M) for the separation of the primary oxalic nucleon. offer the growth access H+(aq) fixation be x⋅mol⋅dm−3.

R C2H2O4(aq)⇌C2HO−4(aq)+H+(aq)

I 0.20

C −x +x

E 0.20−x x

By definition,

Ka1=[C2HO−4(aq)][H+(aq)][C2H2O4(aq)]=5.6⋅10−2

Improving the articulation can provides a quadratic condition regarding x, sinking for x provides [C2HO−4]=x=8.13⋅10−2⋅mol⋅dm−3

(dispose of the negative arrangement since fixations can faithfully be additional noteworthy or appreciate zero).

Presently build a second RICE table, for the separation of the second oxalic nucleon from the amphoteric C2HO−4(aq) particle. offer the modification access C2O2−4(aq) be +y⋅mol⋅dm−3. None of those species was out there within the underlying arrangement. (Kw is neglectable) therefore the underlying centralization of each C2HO−4 and H+ are going to be appreciate that at the harmony position of the first ionization response.

R C2HO−4(aq)⇌C2O2−4(aq)+H+(aq)

I 8.13⋅10−2 zero eight.13⋅10−2

C −y +y

E 8.13⋅10−2−y y eight.13⋅10−2+y

It is smart to just accept that

a. 8.13⋅10−2−y≈8.13⋅10−2,

b. 8.13⋅10−2+y≈8.13⋅10−2 , and

c. The separation of C2HO−4(aq)

Accordingly

Ka2=[C2O2−4(aq)][H+(aq)][C2HO−4(aq)]=1.5⋅10−4≈(8.13⋅10−2 )⋅y8.13⋅10−2

Consequently [C2O2−4(aq)]=y≈Ka2=1.5⋅10−4]⋅mol⋅dm−3

3 0
3 years ago
What is the total number of moles of atoms in one mole of ca3(po4)2?
Westkost [7]
3 Ca 
2 P
8 O

There 13 mols of atoms.
5 0
3 years ago
I need help in balancing chemical equations
Mashcka [7]
I can help. First you want to write out each reactant and product with enough space to write in coefficients. Then box the reactants and products to emphazise how the elements shouldnt change. Afterwards, draw a line in front of the box for you to write in the coefficients.Now we srart balancing. To make things easier, you should always follow the method of : (Metals, Nonmetals,Oxygen,Hydrogen) in that order. By balancing each elememt/chemical this way it should make balancing them easier.
So to acryally perform it you're going to look at your first element (while still following the method) and then look over onto the other side to see what coefficient to place. Then after that element is balanced, you move onto the next element. Remember that everytime you balance an element, start over to make sure all of the other elements are still balanced before proceeding.
3 0
3 years ago
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