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4 years ago

How to balance Na + FeBr3 →

Chemistry

1 answer:

Lubov Fominskaja [6]4 years ago

8 0

Hey there!

Na + FeBr₂ → NaBr + Fe

Firstly, balance Br.

Two on the left, one on the right. Add a coefficient of 2 in front of NaBr.

Na + FeBr₂ → 2NaBr + Fe

Next, balance Na.

One on the left, two on the right. Add a coefficient of 2 in front of Na.

2Na + FeBr₂ → 2NaBr + Fe

Lastly, balance Fe.

One on the left, one on the right. Already balanced.

Our final balanced equation:

2Na + FeBr₂ → 2NaBr + Fe

Hope this helps!

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What is the pH of a 1.0 x 10^-4 M solution of sulfuric acid (H2SO4)?

Marizza181 [45]

Hello!

datos: Molarity = $1.0*10^{-4}\:M\:(mol/L)$

ps: The ionization constant of the sulfuric acid is strong and completely dissociates in water, so the pH will be:

$pH = - log\:[H_3O^+]$

$pH = - log\:[1*10^{-4}]$

$pH = 4 - log\:1$

$pH = 4 - 0$

$\boxed{\boxed{pH = 4}}\end{array}}\qquad\checkmark$

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR!

7 0

3 years ago

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Which element is in group 2 and period 7 of the periodic table

bekas [8.4K]

Hey
Sorry, But There Is No 7th Element in group two. only six which is Radium.
~Spades15

3 0

4 years ago

Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2

Gekata [30.6K]

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide $(CO_2)$ compound

The combination of 2 blue spheres will represent nitrogen molecule $(N_2)$

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = $\frac{2}{2}\times 4=4mol$ of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0

3 years ago

HELP PLEASE!! (100 points)

MrRa [10]

Answer:

As you cool a matter to absolute zero, their kinetic energy reduces significantly and the molecules slows down and begins to aggregate together. ... As heat is added, the molecules gain more kinetic energy. This shown in their increase motion. When heat is withdrawn, the particles slows down hope this helped

3 0

3 years ago

The reform reaction between steam and gaseous methane () produces "synthesis gas," a mixture of carbon monoxide gas and dihydrog

Bond [772]

Answer:

The rate at which dihydrogen gas is being produced = 0.018 kg/s

Explanation:

Firstly, we write the balanced equation for the production of the synthesis gas

CH₄ + H₂O → CO + 3H₂

The rate of consumption of CH₄ is 159 litres per second. With the reaction ran at T = 294°C and a pressure of 0.86 atm

Using the ideal gas equation, we can convert the volumetric rate of consumption of methane to molar rate of consumption

PV = nRT

PV' = n'RT

P = pressure = 0.86 atm = 87,139.5 Pa

V' = 159 L/s = 0.159 m³/s

n' = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 294°C = 567.15 K

87,139.5 × 0.159 = n' × 8.314 × 567.15

n' = (87,139.5 × 0.159) ÷ (8.314 × 567.15)

n' = 2.9383547773 mol/s = 2.938 mol/s

From the stoichiometry of this reaction,

1 mole of methane gives 3 moles of dihydrogen gas

2.938 mol/s of methane will correspond to (3 × 2.938) mol/s of dihydrogen gas, that is, 8.815 mol/s.

Mass flowrate = (molar flowrate) × (molar mass)

Molar flowrate = 8.815 mol/s

Molar mass of dihydrogen gas = 2 g/mol = 0.002 kg/mol

Mass flowrate = 8.815 × 0.002 = 0.0176301287 kg/s = 0.018 kg/s to 2 s.f.

Hope this Helps!!!

3 0

4 years ago

Read 2 more answers