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3 years ago

How many moles of cf4 are there in 171g of cf4 ?

2 answers:

8 0

Convert grams of CF4 to moles by using its molar mass (AKA the masses of all the elements in the compound). Its molar mass is (4(molar mass of F) + 1(molar mass of C)), which is approx. 88.00 grams/mole. Using this, you can set up a proportion or do a quick multiplication using unit analysis, but either way you get about 1.9431 moles.

xeze [42]3 years ago

5 0

1 is + 4 = to n please

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If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 250 mL, what will

kkurt [141]

Answer:B

Explanation:B

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4 years ago

Which natural polymer makes up paper? A. starch B. nylon C. wool D. rubber

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3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

4 years ago

Below are the steps of the engineering process (not necessarily in order): 1) Know the background 2) Do the work 3) Make a plan

LuckyWell [14K]

Answer:

4 - 1 - 3 - 2 - 6 - 5

Explanation:

During an engineering process, first, we need to identify the problem, or the need because the process only will occur because of some need. Then, it's necessary to know as much as possible about the problem and the things that already exist or already were tested to solve it. Knowing the background will make the work easy.

After that, it's necessary to plan the things we'll do, knowing the costs, the time needed for activities, how many people will be necessary for each step, etc. It's really important to make a plan. Then, do the work, following the plan. Thus, the process must be tested. During the test of the results, some problems must be found, so it's time to evaluate and redesign the process, to solve these problems found.

3 0

3 years ago

Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang

Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

$\Delta (H_{f})_{MnO_{2}}$ = -520.0 KJ/mole

$\Delta (H_{f})_{Al_{2}O_{3}}$ = -1699.8 KJ/mole

The balanced chemical reaction is,

$3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)$

Formula used :

$\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}$

$\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))$

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

$\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]$

= (-3399.6) + (1560)

= -1839.6 KJ

5 0

4 years ago