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melamori03 [73]
2 years ago
10

What is a technique for gene therapy?

Chemistry
1 answer:
Maurinko [17]2 years ago
7 0

Answer:

For gene therapy, stem cells can be trained in a lab to become cells that can help fight disease. Liposomes. These fatty particles have the ability to carry the new, therapeutic genes to the target cells and pass the genes into your cells' DNA.

Mark me as brainlest please!

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kvasek [131]

Explanation:

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Al is 2

so is 6

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4 0
2 years ago
What is the atomic number of an atom that has fie neutrons and four electrons
Natali [406]

Answer:

4

Explanation:

3 0
3 years ago
Read 2 more answers
Fe(no3)3 + na0h > fe(oh)3 + nano3 is balanced or unbalanced?
VashaNatasha [74]

Answer:

Unbalanced

Explanation:

\text{Fe}(\text{NO}_3)}_3+ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaNO}_3\\\\\text{Balanced reaction:}\\\\\text{Fe}(\text{NO}_3)}_3+ 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaNO}_3

5 0
2 years ago
18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
Both E. coli and Salmonella are single-celled organisms. They do not have a nucleus or other membrane-bound organelles. Based on
AfilCa [17]

Answer:

domain bacteria

Explanation:

Salmonella and E. coli are same in the sense that they are both bacteria,

7 0
2 years ago
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