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1 year ago

The Lewis structure for a chlorate ion, ClO3-, should show single bond(s), double bond(s), and lone pair(s).

Chemistry

2 answers:

Yuliya22 [10]1 year ago

7 0

Answer: A molecule of $ClO_3^-$ has 1 single bond, 2 double bonds and 8 lone pairs.

Explanation:

Lewis-dot structure is defined as the structure which represents the number of valence electrons around the atoms. The electrons are represented as dots.

This structure helps in the determination of bonding electrons and non-bonding electrons.

For the given compound $ClO_3^-$

As we know that chlorine has '7' valence electrons and oxygen has '6' valence electron. The negative charge gets add up in the total number of valence electrons.

Total number of valence electrons in $ClO_3^-$ = 7 + 3(6) + 1 = 26

In $ClO_3^-$ , two electrons of 2 oxygen atoms combine with 2 electrons of chlorine and 1 electron of one oxygen atom combines with 1 electron of chlorine atom.

A lone pair remains on the central atom which is chlorine.

Hence, a molecule of $ClO_3^-$ has 1 single bond, 2 double bonds and 8 lone pairs.

eduard1 year ago

3 0

1 single bond, 2 double bond 2 lone pairs

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1 year ago

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2 years ago

For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

harina [27]

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

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6 0

2 years ago

I NEED HELP ASAP, WILL MARK BRAINLEST!

Andre45 [30]

Answer:

1. 90%

2. 217.4 g O₂

3. 95.0%

4. Trial 2 ratios

Explanation:

Original: SiCl₄ + O₂ → SiO₂ + Cl₂

Balanced: SiCl₄ + O₂ → SiO₂ + 2Cl₂

Trial SiCl₄ O₂ SiO₂

1 120 g 240 g 38.2 g

2 75 g 50 g 25.2 g

Percentage yield for trial 1

We need to get actual yield (38.2 g) and theoretical yield, in grams.

Mass to moles:

molar mass SiCl₄: 28.09 + 4(35.45) = 169.9 g/mol

120 g SiCl₄ x 1 mol/169.9 g = .706 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.706 mol SiCl₄ = .706 mol SiO₂

Moles to mass:

molar mass SiO₂: 28.09 + 2(16.00) = 60.09 g/mol

.706 mol SiO₂ x 60.09g/mol = 42.44 g SiO₂

Theoretical yield:

actual/theoretical x 100

38.2 / 42.44 = .900 = 90.0% yield

Leftover reactant for trial 1

We know oxygen is the excess reactant.

Mass to moles:

molar mass O₂ = 32.00 g/mol

240 g O₂ x 1 mol/32.00 g = 7.5 mol O₂

We used .706 mol SiO₂, so we also used .706 mol O₂.

7.5 - .706 = 6.8 moles left over

Moles to mass:

6.8 mol O₂ x 32.00g/mol = 217.4 g O₂

Percentage yield for trial 2

Mass to moles:

molar mass SiCl₄: 169.9 g/mol

75 g SiCl₄ x 1 mol/169.9 g = .441 mol SiCl₄

Moles to moles:

For each mole SiCl₄, we have one mol SiO₂ based on the balanced rxn.

.441 mol SiCl₄ = .441 mol SiO₂

Moles to mass:

molar mass SiO₂: 60.09 g/mol

.441 mol SiO₂ x 60.09g/mol = 26.5 g SiO₂

Theoretical yield:

actual/theoretical x 100

25.2 / 26.5 = .950 = 95.0% yield

Because the percentage yield of trial 2 is higher than that of trial 1, we know that the ratio of reactants in trial 2 is more efficient! We got a result closer to our theoretical yield.

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2 years ago

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saw5 [17]

Answer:

Drug calculation

If we have 45g of clobetasol = 0.05%w/w

Then what mass in g of clobetasol is in 0.03%w/w = 45 x 0.03/0.05 =27g

It means that 27g of clobetasol must be added to change the drug strength to 0.03% w/w

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1 year ago

The Lewis structure for a chlorate ion, ClO3-, should show ____ single bond(s), ____ double bond(s), and ____ lone pair(s).

The Lewis structure for a chlorate ion, ClO3-, should show single bond(s), double bond(s), and lone pair(s).