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3 years ago

Which of these is the best definition of biotechnology??

Chemistry

2 answers:

goblinko [34]3 years ago

5 0

C most likely sorry if I’m wrong

Naya [18.7K]3 years ago

4 0

I believe the answer is B) the use and application of living things and science.

Bio means life and technology is scientific knowledge for different purposes.

Sorry if this is wrong

You might be interested in

How many grams of Cl2 must react to produce 0.0923 mol of AlCl3? Show work.

alexandr402 [8]

First, let's state the chemical reaction:

$2Al+3Cl_2\to2AlCl_3\text{.}$

We can find the number of moles of Cl2 required to produce 0.0923 moles of AlCl3, doing a rule of three: 3 moles of Cl2 reacted produces 2 moles of AlCl3:

$\begin{gathered} 3molesCl_2\to2molesAlCl_3 \\ \text{?moles Cl}_2\to0.0923\text{ moles }AlCl_3\text{.} \end{gathered}$

The calculation would be:

$0.0923molesAlCl_3\cdot\frac{3molesCl_2}{2molesAlCl_3}=0.138molesCl_2.$

And the final step is to convert this number of moles to grams. Remember that the molar mass can be calculated using the periodic table, so the molar mass of Cl2 is 70.8 g/mol, and the conversion is:

$0.138molesCl_2\cdot\frac{70.8gCl_2}{1molCl_2}=9.770gCl_2.$

The answer is that we need 9.770 grams of Cl2 to produce 0.0923 moles of AlCl3.

3 0

1 year ago

At the Henry's Law constant for carbon dioxide gas in water is . Calculate the mass in grams of gas that can be dissolved in of

Dvinal [7]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.031M/atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

$C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = $0.0905mol/L$

Volume of solution = 425 mL

Putting values in above equation, we get:

$0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g$

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

8 0

3 years ago

Which scientist is known for developing the planetary model of the atom

xenn [34]

Answer:

Ernest Rutherford

Explanation:

The Rutherford model, proposed originally in 1911, dipicts a planetary model of the atom in which there is a nucleus and electrons circling it.

7 0

3 years ago

Q- Which two positions show the location of the moon in its gibbous phases as seen from Earth?

Phoenix [80]

Answer: 6 & 8

Explanation:

I got it correct on my test hope this helped

4 0

3 years ago

Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h

Gwar [14]

Answer:

0.264 g of $CO_2$ can be formed from 288 mg of $O_2$

Explanation:

The balanced chemical equation is

$2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O$

The conversions are

Mass in mg $O_2$ is converted to mass in g $O_2$

Mass in g $O_2$ is converted to moles $O_2$ by dividing with molar mass

Moles $O_2$ is converted to moles $CO_2$ by using the mole ratio of $O_2:CO_2$ is 9 : 6

Moles $CO_2$ is converted to mass $CO_2$ by multiplying with molar mass $CO_2$

mass in mg $O_2$ > mass in g $O_2$ >moles $O_2$ > moles $CO_2$ > mass $CO_2$

$288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}$

=0.264g (Answer)

7 0

3 years ago