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NikAS [45]
3 years ago
15

In a chemical reaction, which are the most likely products of these reactants 4Fe+302

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
8 0

2Fe2O3, reason is when we add 4Fe + 3O2, we get the same answer, but in a different form.

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How many milliliters of 12.0 m hcl(aq) are needed to prepare 905.0 ml of 1.00 m hcl(aq)?
givi [52]
The  ml  of  12.0 M HCL  needed  to  prepare  905.0 ml  of 1.00m hcl   is  calculated  using   M1V1 =M2V2  formula
M1=  12.0 M
V1=?
M2= 905.0 ml
V2 = 1.00M

V1  is  therefore =  M2V2/M1

=905  x  1.00/12.0 = 75.42 ml

5 0
3 years ago
What term described the movement of around the once a year
gulaghasi [49]
 which describes a term is rotation earths movement around the sun
4 0
3 years ago
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How many moles are represented by 3.01 - 1023 helium atoms?
Pavlova-9 [17]

Answer:

0.5 mol

Explanation:

Given data:

Number of atoms of He = 3.01 ×10²³

Number of moles = ?

Solution:

Avogadro number:  

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.02 × 10²³ is called Avogadro number.

1 mole = 6.02 × 10²³ atoms

3.01 ×10²³ atoms × 1 mol / 6.02 × 10²³ atoms

0.5 mol

7 0
3 years ago
I need help with this
Ganezh [65]
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7 0
3 years ago
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In this reaction, a gaseous system with a volume of 3.00 L has a rate of formation of NOCl of 0.0120 M s-1 . The volume of the s
oee [108]

Answer:

x = 0.324 M s⁻¹

Explanation:

Equation for the reaction can be represented as:

 2 NO(g)   + Cl₂ (g)      ⇄     2NOCl (g)

Rate = K [NO]² [Cl₂]

Concentration = \frac{numbers of mole (n)}{volume (v)}

from the question; their number of moles are constant since the species are quite alike.

As such; if Concentration varies inversely proportional to the volume;

we have: Concentration ∝ \frac{1}{v}

Concentration = \frac{1}{v}

Similarly; the Rate can now be expressed as:

Rate = K [NO]² [Cl₂]

Rate = (\frac{1}{v}) ^2 (\frac{1}{v} )

Rate = \frac{1}{v^3}

We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹

So we can have:

0.0120 = \frac{1}{3^3}

0.0120 = \frac{1}{27}   -----Equation (1)

Now; the new rate of formation when the  volume of the system decreased to 1.00 L can now be calculated as:

x = (\frac{1}{1})^3

x = 1             ------- Equation (2)

Dividing equation (2) with equation (1); we have:

\frac{0.0210}{x} = \frac{\frac{1}{27} }{1}

\frac{0.0210}{x} = \frac{1}{27}

x = 0.0120 × 27

x = 0.324 M s⁻¹

∴  the new rate of formation of NOCl = 0.324 M s⁻¹

8 0
3 years ago
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