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4 years ago

Will give brainliest to the answer!

Chemistry

1 answer:

Nutka1998 [239]4 years ago

7 0

Answer:

a, dipole dipole interactions

Explanation:

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What does C=o represent

Novay_Z [31]

Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is slightly less dense than air. ... Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.

I think it's probably not right.

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3 years ago

What is a Boitic factor

svetoff [14.1K]

Answer:

Biotic components, or biotic factors, can be described as any living component that affects another organism or shapes the ecosystem. This includes both animals that consume other organisms within their ecosystem, and the organism that is being consumed

Explanation:

also you spelt it wrong

4 0

3 years ago

PLEASE HELP!!!

Alisiya [41]

Answer:

1,321.986 (the first one)

6 0

2 years ago

Find the percentage composition of each element in the compound having 9.8 grams of nitrogen,0.7 grams of hydrogen and 33.6 gram

prisoha [69]

Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

Explanation:

Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.

Percentage composition= $\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%$

a) ${\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%$

${\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%$

b) ${\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%$

${\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%$

c) ${\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%$

${\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%$

The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

4 0

3 years ago

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

3 years ago