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bixtya [17]
3 years ago
5

An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was

then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Chemistry
1 answer:
Yuri [45]3 years ago
7 0

Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

<em>Moles HCl added:</em>

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

<em>Moles NaOH to titrate the excess:</em>

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

<em>Moles of acid that were neutralized:</em>

0.0105 moles - 0.0010 moles =

<h3>0.0095 moles of acid were neutralized by the antiacid</h3>
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The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this react
Bond [772]

Answer:

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)

ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln(k₂/k₁)  = 6.92

k₂/k₁ = exp(6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32

3 0
3 years ago
Read 2 more answers
Calculate the molarity of the two solutions.
daser333 [38]

1. 0.33 M

2. 0.278 M

<h3>Further explanation</h3>

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

1. 0.350 mol of NaOH in 1.05 L of solution.

n=0.35

V=1.05 L

Molarity :

\tt M=\dfrac{0.35}{1.05}=0.33

2. 14.3 g of NaCl in 879 mL of solution.

mol NaCl(MW=58.5 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{14.3~g}{58.5~g/mol}=0.244

Molarity :

\tt M=\dfrac{0.244}{0.879~L}\\\\M=0.278

4 0
2 years ago
When naming ionic compounds the ______________ is named first.
grandymaker [24]
The answer is metal. Metals are always named first in ionic compounds, like KNO3 for example. I hope this helps!
7 0
3 years ago
When a 120 g sample of aluminum absorbs 9612 of heat energy, its temperature increases from 25°C to 115°C. Find the specific hea
lesantik [10]
<h3>Answer:</h3>

0.89 J/g°C

<h3>Explanation:</h3>

Concept tested: Quantity of heat

We are given;

  • Mass of the aluminium sample is 120 g
  • Quantity of heat absorbed by aluminium sample is 9612 g
  • Change in temperature, ΔT = 115°C - 25°C

                                                      = 90°C

We are required to calculate the specific heat capacity;

  • We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.

That is;

Q = m × c × ΔT

  • Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.

Specific heat capacity, c = Q ÷ mΔT

                                         = 9612 J ÷ (120 g × 90°C)

                                         = 0.89 J/g°C

Therefore, the specific heat capacity of Aluminium is  0.89 J/g°C

4 0
3 years ago
A gas in a sealed container had its volume increased from 12.1 liters to 21.1 liters.
ArbitrLikvidat [17]

Answer:

The answer to your question is   P2 = 170.9 torr

Explanation:

Data

Volume 1 = 12.1 l                                Volume 2 = 21.1 l

Temperature 1 = 241 °K                      Temperature 2 = 298°K

Pressure 1 = 546 torr                           Pressure 2 = ?

Process

To solve this problem use the combined gas law.

                P1V1/T1 = P2V2/T2

-Solve for P2

                P2 = T1V1T2 / T1V2

-Substitution

                P2 = (241 x 12.1 x 298) / (241 x 21.1)

-Simplification

                P2 = 868997.8 / 5085.1

-Result

                P2 = 170.9 torr

       

8 0
3 years ago
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