If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?
Answer:
›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element
Explanation:
If 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.
What is mass?Mass is defined as a way to gauge how much matter there is in a substance or thing. The kilogram (kg) is the fundamental SI unit of mass, while lower masses can also be measured in grams (g). Atoms make up everyday matter. A majority of an atom's mass is contained in its nucleus.
Given Fe = 38.6 g.
Fe has a molar mass = 55.845 g/mol.
Given mass/molar mass equals 38.6g/55.845gmol-1, or 0.6912 moles of iron.
The reaction is described as Fe + Br2 FeBr2.
One mole Fe yields 1 mole of FeBr2.
FeBr2 would be produced from 0.6912 moles of Fe.
FeBr2 has a molar mass of 215.65 g/mol.
Moles of FeBr2 x Molar mass of FeBr2
= 215.65 g/mole x 0.6912 mole
= 149.06 g FeBr2 produced is the formula.
Thus, if 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.
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Suppose you have samples of three unknown solids. Explain how you could use their properties to
determine whether or not they are ionic solids.
Using melting and boiling temperature, hardness and electric current passing testing.
Ionic solidsIonic solids are materials that have a strong bond between their ions, thus producing well-defined shapes.
In addition, due to this strong attraction, the boiling and melting temperatures of these materials are very high, in addition to the resistance to breakage presented by them.
Finally, ionic solids are also excellent conductors of electricity.
So, their properties used to determine whether or not they are ionic solids are melting and boiling temperature, hardness and electric current passing testing.
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Pleeeeasee someone who’s good at chemistry?! 10 grade
ASAP
I’ll give points, just help please
Answer:
where is the question????????????
3.00 L of a gas is collected at 35.0 C and 0.93 atm. What is the volume at STP
Convert 1.36x10 to standard form
Answer:
13.6 is the correct answer written in standard form.
Explanation:
1.36, move the decimal once to the right to get 13.6
Answer:
13.6
Explanation:
The standard form is 13.6
Thin-layer chromatography explain ?????
Answer:
Explanation:
Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.
Doing Labs at home
I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!
Answer:
go get the stuff.
Explanation:
What identifies the number of protons in the nucleus of an atom?
Answer: Atomic number
Explanation:
I hope this helps you!
Suppose that in an equilibrium mixture of HCl, Cl2, and H2, the concentration of H2 is 1.0 x 10-11 mol-L-1and that of Cl2 is 2.0 x 10-10 mol-L-1. What is the equilibrium molar concentration of HCl at 500 K, given Kc = 4.0 x 1018 for H2(g) +Cl2(g) ⇆ 2HCl(g).
Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].
The balanced reaction is:
H₂(g) +Cl₂(g) ⇆ 2 HCl(g)
Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.
In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.
The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is
[tex]K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }[/tex]
That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case, the constant Kc can be expressed as:
[tex]K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }[/tex]
You know that in an equilibrium mixture of HCl, Cl₂, and H₂:
the concentration of H₂ is 1.0×10⁻¹¹ [tex]\frac{mol}{L}[/tex]the concentration of Cl₂ is 2.0×10⁻¹⁰ [tex]\frac{mol}{L}[/tex]Kc=4×10¹⁸Replacing in the expression for Kc:
[tex]4x10^{18} =\frac{[HCl]^{2} }{[1x10^{-11} ]x[2x10^{-10} ] }[/tex]
Solving:
[tex]4x10^{18} =\frac{[HCl]^{2} }{2x10^{-21} }[/tex]
[tex]4x10^{18} x 2x10^{-21}=[HCl]^{2}[/tex]
[tex]8x10^{-3} =[HCl]^{2}[/tex]
[tex]\sqrt[2]{8x10^{-3}} =[HCl][/tex]
0.0894 [tex]\frac{mol}{L}[/tex]= [HCl]
Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].
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Please help ASAP!!!
Which statement correctly describes the oxidation number of the manganese atom (Mn) in Mnl2 and MnO2?
O Manganese has an oxidation number of +4 in Mnl2 and +2 in MnO2.
o Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
o Manganese has an oxidation number of +4 in both Mnl2 and MnO2.
Manganese has an oxidation number of +2 in both Mnl2 and MnO2.
In this case, according to the given information about the oxidation numbers and the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following [tex]x[/tex], since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:
[tex]Mn^xI_2^-\\\\Mn ^xO_2^{-2}[/tex]
Next, we multiply each anion's oxidation number by the subscript, to obtain the following:
[tex]Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4[/tex]
Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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