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3 years ago

In this reaction, what is the correct coefficient for oxygen gas? ? h2 (g) + ? o2 (g)→ ? h2o(g)

Chemistry

1 answer:

denis-greek [22]3 years ago

7 0

Answer:

1 O₂

Explanation:

In the combustion reaction of hydrogen, oxygen combines with hydrogen to form water as the product of the reaction

Equation of reaction:

2H₂ + O₂ → 2H₂O

From the equation of reaction, it is evident that 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water.

Therefore the coefficient of oxygen in the equation is 1.

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3 years ago

How would i calculate molecules of water produced in a reaction?

levacccp [35]

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4 years ago

Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

2. Volume from density and mass

$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

3. Mass from density and volume

$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water = 12.8 mL

Volume of object = 11.8 mL

$\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}$

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

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3 years ago

What is a double bond? PLEASE HELP

sleet_krkn [62]

Answer:

option c is correct

Explanation:

Hope it helps you......

4 0

3 years ago

Read 2 more answers

What is the value for ∆Soreaction for the following reaction, given the standard entropy values?

Vesnalui [34]

The standard entropy for the substances are as follows:

C6H12O2(s) = -212
O2(g) = -205 
CO2(g) = -214 
H2O(l) = -70

We calculate the ∆S°reaction by the expression:

∆S°rxn = ∆S°products - ∆S° reactants
∆S°rxn = (212+6x205)-(6x214+6x70)
∆S°rxn = -262 J/K ------> OPTION 3

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3 years ago