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3 years ago

12

If some of the gas bubbles from the reaction how to skate out the bottom of the tube, how would this have affected your value fo

r R?

1 answer:

AleksAgata [21]3 years ago

4 0

The gas bubble pops so

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PH que tendra una solución cuya concentracion de [OH-] = 1x10-4

natta225 [31]

I don’t understand wha u just said but thanks for the points

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3 years ago

What is a mineral used in making fertillizer <br>1. silver <br>2. silica <br>3. tin <br>4. gypsum

Marat540 [252]

The Answer Is Gypsum

Hope That Helps!

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4 years ago

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How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl

ratelena [41]

First, we need to determine the half reaction of magnesium. It would be expressed as:

Mg2+ + 2e- = Mg

Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:

4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C

We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.

35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h

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3 years ago

A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas

Anna71 [15]

Answer:

$\large \boxed{\text{528.7 g} }$

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32.00

2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol: 7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

$\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}$

2. Mass of O₂

$\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}$

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3 years ago

How many grams of O are in 675 g of Na2O

PIT_PIT [208]

I think it is .00005815

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3 years ago