What are the positive and negative affects of altering the DNA of a plant cell?

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

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The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

1 year ago

What is true about protons within an atom?

Ira Lisetskai [31]

It defines the element. If you change the protons, you change the type of element. A proton is positively charged and is most of the mass of the atom, next to the neutron. Neutrons have a very very very slightly higher mass.

4 0

3 years ago

A geochemist examines a piece of metal that he found in the soil. He performs tests to identify the metal from its density, elec

aniked [119]

Answer:

what r the statements?

Explanation:

attach or write them

3 0

3 years ago

Help for brainlist and extra points!!

Ratling [72]

Answer:

Explanation:

c.

7 0

3 years ago

Five calcite, CaCO3 (MM 100.085 g/mol), samples of equal mass have a total mass of 12.2±0.1 g . What is the average mass and abs

Westkost [7]

Answer:

The average mass of calcium in each sample is: 0.978 g

The absolute uncertainty is: 0.008 g

Explanation:

The absolute uncertainty of the total samples indicated in the statement is ± 0.1 g.

When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the relative uncertainties together.

The relative uncertainty is the absolute uncertainty divided by the quantity:

Relative uncertainty = 0.1g / 12.2 g = 0.008

The average mass of calcium is calculated using proportions, along with the molar masses:

Molar mass of calcium: 40.078 g/ mol (from a periodic table)

Molar mass of calcite: 100.085 g/mol (given)

Proportion:

40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite

x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium

So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:

Average mass = total mass of five samples / number of samples

Average mass = 4.89 g / 5 = 0.978 g of calcium

So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).

The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:

Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g

The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.

7 0

4 years ago