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3 years ago

Perform the following

Chemistry

1 answer:

goblinko [34]3 years ago

5 0

Answer:

8.9 or 8.979

Explanation:

When you add or subtract, you assign significant figures in the answer based on the number of decimal places in each original measurement.

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3 years ago

15 mL of a 0.16 M solution of glucose was diluted until the total volume reached 95 mL. What is the concentration of new solutio

Nana76 [90]

Answer:

the concentration of new solution = 0.025 M

Explanation:

Given that :

The initial volume V_1 = 15 mL

Initial concentration M_1 = 0.16 M

Final volume V_2 = 95 mL

Final concentration M_2 = ???

We know that M_1 × V_1 = M_2V_2

Making M_2 the subject of the formulae; we have :

M_2 = (M_1 × V_1)/V_2

M_2 = (0.16 × 15)/ 95

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How many valence electrons does zinc oxide have?

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3 years ago

Read 2 more answers

I need help on both a and b of question 1

marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

$r = \frac{\Delta c}{\Delta t}$

Given the following reaction:

$2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)$

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

$r = -\frac{\Delta [N_2O_5]}{2 \Delta t}$

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

$r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}$

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

$r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s$

(b) Using the relationship derived previously, we know that:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

Rate of appearance of nitrogen dioxide is given by:

$r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}$

Which is obtained from the equation:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

If we multiply both sides by 4, that is:

$-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}$

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

5 0

3 years ago