CHEMISTRY COLLEGE

The amount of I − 3 ( aq ) in a solution can be determined by titration with a solution containing a known concentration of S 2 O 2 − 3 ( aq ) (thiosulfate ion). The determination is based on the net ionic equation 2 S 2 O 2 − 3 ( aq ) + I − 3 ( aq ) ⟶ S 4 O 2 − 6 ( aq ) + 3 I − ( aq ) Given that it requires 29.4 mL of 0.380 M Na 2 S 2 O 3 ( aq ) to titrate a 30.0 mL sample of I − 3 ( aq ) , calculate the molarity of I − 3 ( aq ) in the solution.

Answers

Answer 1

Answer:

Answer:

The molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

Explanation:

Given net ionic equation:

2S₂O₃²⁻ (aq) + I₃⁻ ( aq ) ⟶ S₄O₆²⁻ (aq) + 3I⁻ (aq)

Number of moles of S₂O₃²⁻: n₁ = 2, Number of moles of I₃⁻: n₂ = 1

Given- For S₂O₃²⁻ solution: Molarity: M₁ = 0.380 M, Volume: V₁ = 29.4 mL;

For I₃⁻ (aq) solution: Molarity: M₂ = ? M, Volume: V₂ = 30.0 mL

To calculate the molarity of I₃⁻ (aq) solution, we use the equation:

$(M_(1)V_(1))/(n_(1))=(M_(2)V_(2))/(n_(2))$

$((0.380 M)* (29.4 mL))/(2)=(M_(2)* (30.0 mL))/(1)$

$\Rightarrow M_(2) = ((0.380 M)* (29.4 mL))/((30.0 mL)* 2) = 0.186 M$

Therefore, the molarity of I₃⁻ (aq) solution: M₂ = 0.186 M

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Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make?

Answers

Answer:- 353.3 g

Solution:- The balanced equation is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

We start with given grams of calcium chloride and convert them to moles. Then using mol ratio, the moles of calcium phosphate are calculated and converted to grams as.

Molar mass of calcium chloride is 110.98 gram per mol and molar mass of calcium phosphate is 310 gram per mol.

The set is made using dimensional analysis as:

$379.4gCaCl_2((1molCaCl_2)/(110.98gCaCl_2))((1molCa_3(PO_4)_2)/(3molCaCl_2))((310gCa_3(PO_4)_2)/(1molCa_3(PO_4)_2))$

= $353.3gCa_3(PO_4)_2$

So, 353.3 grams of calcium phosphate can be formed.

Answer:

353.3g

Explanation:

What will the pressure of the gas be if the temperature rises to 87°C?A. 60 kPa
B. 161 kPa
C. 16 kPa
D. 41 kPa

Answers

Answer:

A. 60 kPa

Explanation:

P2 = p1 times t2 / t1

Why can't 1−methylcyclohexanol be prepared from a carbonyl compound by reduction? select the single best answer?

Answers

1−methylcyclohexanol is a tertiary alcohol. Tertiary Alcohols are synthesized by either reacting Ketone with Organometallic compounds like Grignard reagent or by hydration of substituted alkenes. 1−methylcyclohexanol can not be synthesized by reduction of carbonyl compound because it is not possible to have a starting carbonyl compound having carbonyl group along with three other alkyl groups (as carbon can only form 4 bonds).

Result:
Tertiary alcohols don't contain a hydrogen atom at carbon attached to hydroxyl group that is why it is not possible to synthesize 1−methylcyclohexanol by reduction of carbonyl compound.

is the number of total atoms on the left side of a balanced equation always equal to the number of total atoms on the right side of the equation?

Answers

Yes, because conservation of mass

Yes each and every timee,

Which of the following forces causes comets to regularly return to the inner solar system after being gone for many years?A. friction
B. magnetism
C. electromagnetic force
D. gravity

Answers

Answer is D. Gravity

Answer:

✨Gravity✨ hope this helps

Explanation:

The federal limit for cadmium in drinking water is 0.010 mg per liter of solution. What is the molar concentration of a Cd solution that has reached the limit?

Answers

Thank you for posting your question here. The molar concentration of a Cd solution that has reached the limit is 8.89x10^-8 mols/L. The equation to be used M=n/L to solve the above problem. Below is the solution:

.010mg/L = .00001g/L
.00001g / 112.41g/mol = 8.89x10^-8 mols/L

Answer:

$M=8.9x10^(-8)M$

Explanation:

Hello,

In this case, one can assume 1L as the volume of the solution, so we've got 0.010mg of cadmium. Now, as we're asked to know its molarity, one computes the moles of cadmium as follows:

$n_(Cd)=0.010mg*(1x10^(-3)gCd)/(1mgCd)*(1molCd)/(112.4gCd)=8.9x10^(-8)molCd$

Now, one obtains the molar concentration (molarity) as shown below: $M=(8.9x10^(-8)molCd)/(1L)\n M=8.9x10^(-8)M$

Best regards.

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