How many moles would needed to be dissolved in 1 dm3 to get a 3M solution? How many would be needed if only 0.1 dm3 were used?

Why are carbon and nitrogen essential for survival to all living organisms? A.

bagirrra123 [75]

Hello there.

<span>Why are carbon and nitrogen essential for survival to all living organisms?
</span>
<span>They are needed to release the energy stored in food.
</span>Therefore, The correct answer would be Option B.

3 0

3 years ago

Read 2 more answers

Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air

Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s$

b) for completion of 32% of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-32}$

$t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}$

$t=44.3s$

c) for completion of 81 % of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-81}$

$t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}$

$t=191s$

4 0

2 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

(4.9 x 10-2) (9.80 x 102) =

Viktor [21]

The answer is 46981.2

5 0

2 years ago

What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso

Paraphin [41]

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

6 0

3 years ago