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3 years ago

Help please I’m stuck on thsi question

Chemistry

2 answers:

Ira Lisetskai [31]3 years ago

8 0

The answer is B and I have to add this so I can submit the answer

vodomira [7]3 years ago

7 0

Since it’s asking for chemical to thermal, B.

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What is the molar mass of Na2CO3? 60.0 g/mol 106.0 g/mol 118.0 g/mol 141.0 g/mol

Dmitry [639]

Answer is: the molar mass od sodium carbonate (Na₂CO₃) is 106.0 g/mol.

M(Na₂CO₃) = 2 · Ar(Na) + Ar(C) + 3 · Ar(O).

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M(Na₂CO₃) = 46 + 12 + 48 · g/mol.

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The Electrical conductivity of an aqueous solution depends on the concentration of which particles in the solution

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3 years ago

Read 2 more answers

Which of the following changes would improve the accuracy of the molar mass measurement carried out in this experiment:

aleksandrvk [35]

A. would not help and maybe make it harder to get accurate measurements.

b. would make more delta T which will improve its accuracy

c. will also increase delta T.

4 0

3 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.

6 0

3 years ago