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DerKrebs [107]
3 years ago
13

Sulfur powder (S) and oxygen gas (O2) undergo a

Chemistry
2 answers:
Zigmanuir [339]3 years ago
5 0

S (s) + O₂ (g) → SO₂ (aq) represents this chemical reaction

<h3>Further explanation</h3>

In stating a chemical equation it can be done in the form of a word or a chemical formula

A word equation will include the words of the reactants, products, form of the compound (liquid, gas, solid), the total concentration/quantity of the reactants and products which can be expressed in mass, moles, or volume

The questions statement above shows that sulfur powder (S) and oxygen gas (O₂) are reactants (located on the left of the reaction equation), and sulfur dioxide (SO₂) is the product of the reaction (stated in the problem there is the word "to form") which is located on the right

So the complete reaction

S (s) + O₂ (g) → SO₂ (aq)

MaRussiya [10]3 years ago
4 0

Answer:

The correct answer is A

Explanation:

You might be interested in
10 + 32 -6=any one?​
Kryger [21]

Answer:

Your answer will be 36

Explanation:

10+32=42

42-6=36

3 0
3 years ago
What happens to the liquid in a thermometer when it is moved from cold water to boiling water?
MissTica
Nothing it just changes temp as it is protected by glass layers
5 0
3 years ago
Read 2 more answers
How many grams of O are in 635 g of Li 2 O ?
KiRa [710]

Answer:

340g

Explanation:

Lithium oxide or Li2O is an inorganic compound made of two lithiums and one oxygen molecules. Lithium molecular mass is 6.94g/mol while oxygen molecular mass is 16g/mol. The molecular mass of Li2O will be:  

2* 6.94g/mol + 1*16g/mol= 29.88g/mol

Out of 1 mol lithium oxide (29.88g), there is 1 mol of oxygen(16g). Then, out of 635g lithium oxide the number of oxygen will be: 635g * (16g/29.88g)= 340g

5 0
3 years ago
16.25 g of water at 54 C relaeases 402.7 J. What will be its final temp?
leonid [27]
Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?

Solving:

Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
\Delta\:T =  \frac{-402.7}{67.99}
\boxed{\Delta\:T \approx -5.92\:^0C}

If: ΔT (T final - T initial) = ?
-5.92^0 =  T_{final} -  54^0
T_{final} = 54^0 - 5.92^0
\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark

8 0
4 years ago
54.56 g of water at 80.4 oC is added to a calorimeter that contains 47.24 g of water at 40 oC. If the final temperature of the s
fomenos

Answer:

49.5J/°C

Explanation:

The hot water lost some energy that is gained for cold water and the calorimeter.

The equation is:

Q(Hot water) = Q(Cold water) + Q(Calorimeter)

<em>Where:</em>

Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J

Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J

That means the heat gained by the calorimeter is

Q(Calorimeter) = 4794J - 3834J = 960J

The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:

59.4°C-40°C = 19.4°C

And calorimeter constant is:

960J/19.4°C =

<h3>49.5J/°C</h3>

<em />

7 0
3 years ago
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