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Katarina [22]
3 years ago
12

Why is the formation of water a synthesis reaction?

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
7 0

Explanation:

A synthesis reaction: It is defined as a kind of reaction where one and more than one reactant attached and creates an individual product.

The formation of the water is an example of a synthesis reaction because here more than one reactants combine and create a single product (water). Water formation occurs when 2 hydrogens and an oxygen share electrons through covalent bonds.

2H + O ----> H2O.

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A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r
aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:

  • V₁n₂=V₂n₁

Where in this case:

  • V₁ = 6.13 L
  • n₂ = ?
  • V₂ = 11.3 L
  • n₁ = 2.51 mol

We <u>input the data</u>:

  • 6.13 L * n₂ = 11.3 L * 2.51 mol
  • n₂ = 4.63

As <em>4.63 moles is the final number of moles</em>, the number of moles added is:

  • 4.63 - 2.51 = 2.12 moles
5 0
2 years ago
If the pressure inside the cylinder increases to 1.3 atm, what is the final
EleoNora [17]

Answer:

1.4 × 10² mL

Explanation:

There is some info missing. I looked at the question online.

<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>

Step 1: Given data

  • Initial volume (V₁): 215 mL
  • Initial pressure (P₁): 625 mmHg
  • Final volume (V₂): ?
  • Final pressure (P₂): 1.3 atm

Step 2: Convert 625 mmHg to atm

We will use the conversion factor 1 atm = 760 mmHg.

625 mmHg × 1 atm/760 mmHg = 0.822 atm

Step 3: Calculate the final volume of the air

Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL

5 0
3 years ago
In our simulation what did the σ (sigma) stand for?
soldier1979 [14.2K]

Answer:

C -The atomic radius

Explanation:

I took the test :)

6 0
3 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
What is the molecular formula of a compound containing 42.17% Chlorine, 57.83% Oxygen, and has a molecular mass of 166 g/mol?
Juliette [100K]

Answer:

C2H4CL2

Explanation:

this is your answer

6 0
3 years ago
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