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2 years ago

What does two bonded flurione atoms form?

1 answer:

8 0

2 Fluorine atoms covalently bonded with each other, each sharing an equal number of a single, one valence electron to achieve a stable octet, would form a diatonic fluorine gas. It is a diatomic molecule, a molecule consisting of 2 atoms that are the same, in this case fluorine.

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I need help solving a kinetics equation.

Lady bird [3.3K]

Answer:

what would the question be?

7 0

2 years ago

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

VMariaS [17]

M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C

m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C

T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C

7 0

3 years ago

A first order reaction B=C has a half life of

Lostsunrise [7]

Answer:

90.46 well that's how I got the answer from my calculations

5 0

2 years ago

Which or these equations is balanced H2SO4 + 2Al > Al2(SO4)3 + H2 or 2KCl + Pb(NO3)2 > 2KNO3 + PbCl2

Dmitry_Shevchenko [17]

Balancing of chemical equation is essential because of the law of conservation of mass, which states that the mass of a system can not be created or removed.

The second equation is balanced

$2KCl_(_a_q_) +Pb(NO_3)_2_(_a_q_) ==> 2KNO_3_(-a_q_) + PbCl_2(_a_q_)$

This is because the number of elements of each atom in the product side equal the number of elements of each atom on the reactant side.

The first equation is not balanced

$H_2SO_4 _(_a_q) + 2Al_(_s_) ==> Al(SO_4)_3_(_a_q_) + H_2_(_g_)$

This is because there is 1 molecule of $SO_4^{-2}$ on reactant side as compared to 3 molecules of $SO_4^{-2}$

To balance the equation we add a coefficient of 3 on sulphuric acid ( $H_2SO_4$ ) and a coefficient of 3 on hydrogen ( $H_2$ )

$3H_2SO_4_(_a_q_) + 2Al_(_s_) ==> Al(SO_4)_3_(_a_q_) + 3H_2_(_g_)$

7 0

3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago