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Pavel [41]
2 years ago
11

During a lab exam a student looked through the microscope and identified the following cell parts: circular DNA, ribosomes, cell

wall, and cytoplasm. What type of cell did she identify?
Eukaryotic

Endoplasmic

Prokaryotic

None of the above
Chemistry
1 answer:
notsponge [240]2 years ago
5 0

Answer:

The correct answer to this question is a prokaryotic cell.

Explanation:

Prokaryotes have circular DNA and a cell wall, whereas eukaryotes have their DNA packaged into chromosomes and may not have a cell wall (as is the case in animal cells).

Hope this helps!

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Rank these acids according to their expected pKa values.ClCH2COOHClCH2CH2COOHCH3CH2COOHCl2CHCOOHIn order of highest pka to lowes
Ilia_Sergeevich [38]

Answer:

CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH  > ClCH₂COOH

Explanation:

Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.

Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.

  • The closer the substituent is to the carboxyl group, the greater is its effect.
  • The more substituents, the greater the effect.  
  • The effect tails off rapidly and is almost zero after about three C-C bonds.

CH₃CH₂-CH₂COOH —  EDG —                         weakest —  pKₐ = 4.82

      CH₃-CH₂COOH — reference —                                     pKₐ = 4.75

  ClCH₂-CH₂COOH — EWG on β-carbon— stronger —     pKₐ = 4.00

           ClCH₂COOH — EWG on α-carbon — strongest —  pKₐ = 2.87

6 0
2 years ago
Read 2 more answers
An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi
Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow  = 713.51 N/m

3 0
2 years ago
Joebert had a rectangular cup and filled it with water. He measured the water's mass and volume. Then he got a density of 1.25g/
ANTONII [103]

Answer:

Percent error = 25%

Explanation:

Given data:

Measured density of water = 1.25 g/mL

Accepted density value of water = 1 g/mL

Percent error = ?

Solution:

Formula:

Percent error = (measured value - accepted value / accepted value) × 100

Now we will put the values in formula:

Percent error = (1.25 g/mL - 1 g/mL /1 g/mL )× 100

Percent error = (0.25 g/mL /1 g/mL )× 100

Percent error = 0.25 × 100

Percent error = 25%

5 0
3 years ago
describe how a pure dry sample of solid lead carbonate can be obtained from sodium carbonate solution and lead nitrate solution
djyliett [7]

Answer:

Soluble salts can be made by reacting acids with soluble or insoluble reactants. Titration must be used if the reactants are soluble. Insoluble salts are made by precipitation reactions.

Making insoluble salts

An insoluble salt can be prepared by reacting two suitable solutions together to form a precipitate.

Determining suitable solutions

All nitrates and all sodium salts are soluble. This means a given precipitate XY can be produced by mixing together solutions of:

X nitrate

sodium Y

For example, to prepare a precipitate of calcium carbonate:

X = calcium and Y = carbonate

mix calcium nitrate solution and sodium carbonate solution together

calcium nitrate + sodium carbonate → sodium nitrate + calcium carbonate

Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3(aq) + CaCO3(s)

It also works if potassium carbonate solution or ammonium carbonate solution is used instead of sodium carbonate solution. Remember that all common potassium and ammonium salts are soluble.

please mark as brainliest

Explanation:

7 0
2 years ago
A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th
tatyana61 [14]

Answer : The partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % CO_2 and 25 % N_2 in terms of volume.

First we have to calculate the moles of CO_2 and N_2.

\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole

\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole

Now we have to calculate the mole fraction of CO_2.

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}

\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75

Now we have to calculate the partial pressure of the CO_2 gas.

\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}

\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

3 0
2 years ago
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