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3 years ago

What is a family in a periodic table?

Chemistry

1 answer:

igomit [66]3 years ago

8 0

Answer:

A family is a vertical column .

The elements in a family have similar chemical properties.

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A theory is:

SpyIntel [72]

A theory is a proposed explanation for an observation

3 0

2 years ago

What is the limiting reactant in a reaction where 10.0 mol of iron is treated with 12.0 mol of bromine? The product that forms i

hammer [34]

Answer: The limiting reagent in the reaction is bromine.

Explanation:

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.

Given values:

Moles of iron = 10.0 moles

Moles of bromine = 12.0 moles

The chemical equation for the reaction of iron and bromine follows:

$2Fe+3Br_2\rightarrow 2FeBr_3$

By the stoichiometry of the reaction:

If 3 moles of bromine reacts with 2 moles of iron

So, 12.0 moles of bromine will react with = $\frac{2}{3}\times 12.0=8moles$ of iron

As the given amount of iron is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Hence, bromine is considered a limiting reagent because it limits the formation of the product.

Thus, the limiting reagent in the reaction is bromine.

3 0

3 years ago

را

Aloiza [94]

Answer:

A.It is the same for every sample of a single substance.

Explanation:

7 0

3 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago

A sample of gas (1.9 mol) is in a flask at 21 °c and 697 mm hg. the flask is opened and more gas is added to the flask. the new

garik1379 [7]

To solve this problem, we assume ideal gas so that we can use the formula:

PV = nRT

since the volume of the flask is constant and R is universal gas constant, so we can say:

n1 T1 / P1 = n2 T2 / P2

1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) / 841 mm Hg

n2 = 2.25 moles

8 0

3 years ago