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Aneli [31]
3 years ago
15

If the reactants of a reaction have one s and four o atoms, what will the products have?

Chemistry
2 answers:
Anton [14]3 years ago
6 0
Answer:
            The products will also have one S and four O atoms.

Explanation:
                   According to Law of Conservation of Matter, " Matter can neither be created nor destroyed, but it can be changed from one form into another".
                   So, in given scenario the number of atoms in reactants and products will also remain the same, but their chemical composition can vary. They might have no similarity in chemical composition but their numbers will remain the same.
                   The reaction might be the decomposition of sulfuric acid,, 

                                        H₂SO₄     →     SO₃  +  H₂O

In above case number of S and O on both sides are same, but chemical composition has changed.
alexandr402 [8]3 years ago
4 0

Answer:

one S four O atoms

Explanation:

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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
ddd [48]

Answer:

178.35g

Explanation:

Molarity of a solution can be calculated using the formula:

Molarity = number of moles ÷ volume

Based on the information provided in this question, molarity (M) of the solution = 1.50 M, volume = 725 mL = 725/1000 = 0.725L, n = ?

1.50 = n / 0.725

n = 1.50 × 0.725

n = 1.0875mol

Molar mass of Na3PO4

23(3) + 31 + 16(4)

= 69 + 31 + 64

= 164g/mol

Mole = mass ÷ molar mass

1.0875 = mass/164

mass = 178.35g

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Explanation:

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An aqueous solution of potassium sulfate is allowed to react with an aqueous solution of calcium nitrate. identify the solid in
Maksim231197 [3]
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Which item has the most thermal energy?
Ksju [112]

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Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0
3 years ago
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