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3 years ago

14

Based on the current atomic theory most chemical reaction take place as a result of the interactions of which sub- atomic partic

les

1 answer:

icang [17]3 years ago

6 0

Electrons, everything is pretty much based around the likeliness of electrons to be swapped or shared between atoms

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What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?

Aleksandr [31]

Answer:

$\boxed {\boxed {\sf 3.3 \ liters}}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}$

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

molarity= 1.2 mol/L
moles of solute= 4.0 mol
liters of solution =x

Substitute the values into the formula.

$1.2 \ mol/L = \frac{4.0 \ mol}{x}$

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

$\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}$

$4.0 \ mol *1=1.2 \ mol/L *x$

$4.0 \ mol = 1.2 \ mol/L *x$

x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L

$\frac{4.0 \ mol}{1.2 \ mol/L} = \frac{1.2 \ mol/L *x}{1.2 \ mol/L}$

$\frac{4.0 \ mol}{1.2 \ mol/L}=x$

The units of moles (mol) will cancel.

$\frac{4.0 }{1.2 } \ L =x$

$3.33333333 \ L=x$

The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.

The 3 in the hundredth place tells us to leave the 3 in the tenths place.

$3.3 \ L\approx x$

Approximately <u>3.3 liters of solution</u> are needed to make a 1.2 M solution with 4.0 moles of sodium chloride.

7 0

3 years ago

Complete the following radioactive decay problem.

zzz [600]

I think it’s the third option

7 0

3 years ago

Read 2 more answers

Which statements correctly describe the decay rates of radioactive isotopes?

storchak [24]

Answer: b} The exact time when an individual atom will decay can be accurately predicted.

c} After each half-life, the amount of radioactive material is reduced by half.

Explanation:

All radioactive decay follows first order kinetics.

Rate law expression for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant

a - x = amount left after decay process

Expression for calculating half life, which is the time taken by the half of the reactants to decompose is:

$t_{1/2}=\frac{0.693}{k}$

4 0

3 years ago

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Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$

7 0

3 years ago

Can someone please help me with this?

yaroslaw [1]

Answer:

Lonic.an electron will be transferred from potassium to the chlorine atom

3 0

3 years ago

Read 2 more answers