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abruzzese [7]
2 years ago
8

A scientist wants to create a drug that will kill cancer cells upon absorption in the intestines. In order for the drug to work

effectively, it must not be absorbed until it reaches the intestines. The scientist tests different acidities of compounds to determine which combination will keep the drug intact until it reaches the intestines. What part of the engineering process is the scientist engaged in?
Chemistry
2 answers:
Anarel [89]2 years ago
8 0
The testing of the acidities of the compounds to determine the combination that will keep the drug intact before it reaches the intestines is a part of EXPERIMENTATION process. The experimentation is done in order to prove or answer different queries that were raised during the hypothesis making. 
OlgaM077 [116]2 years ago
7 0

<u><em>make a plan</em></u> i

s the answer for my apex users

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A person is pushing a box across a table. The measured forces on the box are 10N, 14N, 7N, 10N. Which force is represented by 7N
Vlad1618 [11]
Answer: The friction force.

Justification:

Since the box is sliding over the table, the normal force equals the weight of the object (and any other vertical force that is applied on the box).

So, the normal force and weight (along with any other vertical component of a force applied on the box) must be 10 N and 10 N.

The other two forces: 14 N and 7 N are the forces in the plane of the table and should be opposite in a same line. The 14 N force is the responsible of the motion and the 7N force is opposing the 14 N force, so the 7N force has to be the friction force. Of course, 14N - 7N > 0 which is why the box is moving.
3 0
2 years ago
Read 2 more answers
What is the mole fraction of potassium dichromate, K2Cr2O7, in a solution prepared from 24.42 g of potassium dichromate and 240.
Harrizon [31]

Answer:

6.19\times 10^{-3} is the mole fraction of potassium dichromate.

Explanation:

Mass of potassium dichromate = 24.42 g

Moles of  potassium dichromate =n_1=\frac{24.42 g}{294.185 g/mol}=0.0830 mol

Mass of water = 240.0 g

Moles of water =n_2=\frac{240.0 g}{18.015 g/mol}=13.3222 mol

Mole fraction is calculated by:

\chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{0.0830 mol}{0.0830 mol+13.3222 mol}=0.00619=6.19\times 10^{-3}

6.19\times 10^{-3} is the mole fraction of potassium dichromate.

8 0
2 years ago
Which products would result from the double replacement reaction between mgcl2(aq) and na2co3(aq)?
Andrew [12]
Double displacement reactions are when the cations and anions of 2 compounds are exchanged. Cation of one reactant will form a new product with the anion of the other reactant and vice versa.
MgCl₂ --> Mg²⁺  + 2Cl⁻
Na₂CO₃ --> 2Na⁺ + CO₃²⁻
the cations and anions are exchanged therefore double displacement reaction is as follows;
MgCl₂(aq) + Na₂CO₃(aq) --> MgCO₃(s) + 2NaCl(aq)
8 0
1 year ago
The student collects the H2(g) produced by the reaction and measures its volume over water at 298 K after carefully equalizing t
lubasha [3.4K]

Answer:

The pressure of  H₂(g) = 741 torr

Explanation:

Given that:

The atmospheric pressure measured in the lab  = 765 torr

The vapor pressure of water = 24 torr

By applying Dalton's Law of Partial Pressure :

P_{total} = P_{H_2}+P_{H_2O}

Making The Pressure inside the tube due to the H₂(g) the subject of the formula :

we have:

P_{H_2} = P_{total} -  P_{H_2O}

= (765 -24) torr

= 741 torr

Therefore; the pressure of  H₂(g) = 741 torr

4 0
2 years ago
For the reaction, 2cr2+ + cl2(g) ---&gt; 2cr3+ + 2cl- e cell (standard conditions) = 1.78v calculate ecell (standard conditions)
PolarNik [594]

Answer:

-1.78 V

Explanation:

There are several rules required to calculate the cell potential:

  • given standard cell potential, we may reverse the equation: the products of a given reaction become our reactants, while reactants become our products in the reversed equation. For a reversed equation, we change the sign of the cell potential to the opposite sign;
  • if we multiply the whole equation by some number, this doesn't influence the cell potential value. It only produces a different expression in the equilibrium constant.

That said, notice that the initial reaction with respect to the final reaction is:

  • reversed: chromium(III) cation and chloride anion become our reactants as opposed to the products in the initial reaction, so we change the sign of the cell potential to a negative value of -1.78 V;
  • each coefficient is multiplied by a fraction of \frac{1}{2}. It doesn't influence the value of the cell potential.

Thus, we have a cell of E = -1.78 V.

3 0
2 years ago
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