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UNO [17]
3 years ago
8

Help asap ill give branilsit

Chemistry
1 answer:
stiks02 [169]3 years ago
3 0
A prokaryotic cell is a cell without a nucleus (there's more to it, though).

Let's start from the bottom.

Strawberry plant: Plant cells are eukaryotic (cells have nucleus).
Hippopotamus: Animal cells are eukaryotic (cells have nucleus).
Paramecium: That's an animal, and animal cells are eukaryotic (cells have nucleus).
Bacterium: Perfect! Bacterium are prokaryotes! (cells don't have nucleus).

Your answer is bacterium.

Have an awesome day! :)
You might be interested in
1 Aluminum reacts with oxygen to produce aluminum oxide, what is the chemical equation?​
Roman55 [17]

Answer:  Aluminum reacts with oxygen to produce aluminum oxide as follows: 4Al + 3O2 → 2Al2O3 a.

Explanation: I really really <u>hope</u> that was <u>helpful.</u>

4 0
3 years ago
What is the molality of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175? (The molar mass of KCl = 74.55 g/mol
NARA [144]

Answer:

The molality of the KCl solution is 11.8 molal

Explanation:

Step 1: Data given

Mol fracrion KCl = 0.175

Molar mass KCl = 74.55 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mol fraction H2O

mol fraction H2O = 1 - 0.175 = 0.825

Step 3: Calulate mass of H2O

Suppose the total moles = 1.0 mol

Mass H2O = moles H2O * molar mass

Mass H2O = 0.825 * 18.02 g/mol

Mass H2O = 14.87 grams = 0.01487 kg

Step 4: Calculate molality

Molality KCl = 0.175 / 0.01487 kg

Molality KCl = 11.8 molal

The molality of the KCl solution is 11.8 molal

5 0
3 years ago
A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

Next, the total moles:

n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0
3 years ago
What volume would a sample of gas occupy in LITERS at 0.985 atmospheres and a volume of 3.65 liters if the pressure were raised
musickatia [10]

Answer:

3.18 L

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 0.985 atm
  • Initial volume (V₁): 3.65 L
  • Final pressure (P₂): 861.0 mmHg
  • Final volume (V₂): ?

Step 2: Convert P₁ to mmHg

We will use the conversion factor 1 atm = 760 mmHg.

0.985 atm × 760 mmHg/1 atm = 749 mmHg

Step 3: Calculate the final volume of the gas

Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L

7 0
3 years ago
What is the density of a gas with a volume of 5.6 FL oz. and a mass of 37 grams?
Assoli18 [71]

Answer:

6.60

Explanation:

4 0
3 years ago
Read 2 more answers
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