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3 years ago

Is N2 + H2 > NH3 balanced?

Chemistry

1 answer:

polet [3.4K]3 years ago

4 0

Answer:

yes

Explanation:

This can be fixed by multiplying the product contacting nitrogen by 2. o The new chemical equation is N2 +H2 →2 NH3 Reactants Products Nitrogen 2 2 Hydrogen 2 6 Since NH3 is multiplied by a coefficient of 2 there are now 2 nitrogens and 6 hydrogens. The 6 hydrogens come from the 2 multiplied by the subscript of 3.

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If 45.0 mL of a 0.0500 M HNO3, 10.0 mL of a 0.0500 M KSCN, and 30.0 mL of a 0.0500 M Fe(NO3)3 are combined, what is the initial

jeka94

Answer:

the initial concentration of SCN- in the mixture is 0.00588 M

Explanation:

The computation of the initial concentration of the SCN^- in the mixture is as follows:

As we know that

$KSCN \rightarrow K^ + SCN^-$

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M

So, the total milimoles of SCN^- is

= 10 × 0.05

= 0.5 m moles

The total volume in mixture is

= 45 + 10 + 30

= 85 mL

Now the initial concentration of the SCN^- is

= 0.5 ÷ 85

= 0.00588 M

hence, the initial concentration of SCN- in the mixture is 0.00588 M

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3 years ago

What could we use to measure the energy of particles when we are unable to see them

iVinArrow [24]

Answer:

Temperature.

Explanation:

Temperature is a measure of the average kinetic energy of a system.

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3 years ago

5) O etanol C2H6O, combustível produzido em grande escala no Brasil, pode ser obtido pela fermentação da sacarose C12H22O11 enco

UNO [17]

Answer:

a) 0.15

Explanation:

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3 years ago

You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1

jeka57 [31]

Answer:

$\boxed{\text{36 mL}}$

Explanation:

1. Write the balanced chemical equation.

$\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O$

2. Calculate the moles of HCOOH

$\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}$

3. Calculate the moles of NaOH.

$\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}$

4. Calculate the volume of NaOH

$c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}$

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3 years ago

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ivann1987 [24]

Answer:

its on goggle!

Explanation:

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3 years ago