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2 years ago

These type of cells lack a nucleus -

Chemistry

2 answers:

nexus9112 [7]2 years ago

5 0

Prokaryotes are organisms whose cells lack a nucleus and other organelles

lesya [120]2 years ago

3 0

Answer:

Prokaryotes are cells that lacks nucleus

Explanation:

Hope you have a great amazing day and winter break:)

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PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro

podryga [215]

Answer :

Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part 2 : The theoretical yield of $NH_3$ gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

<u>Answer for Part (2) :</u>

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

3 0

3 years ago

Mechanical energy is the sum of the _____ energy of an object. A kinetic and potential B potential and thermal C kinetic and che

Juli2301 [7.4K]

Answer:

A. kinetic and potential

Explanation:

kinetic energy (energy of motion) or potential energy (stored energy of position)

- Hope that helped! Let me know if you need further explanation.

5 0

2 years ago

How many moles are in 3.01 x 10^23 atoms of zinc?

Lerok [7]

Answer:

<h2>0.5 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.01 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{3.01}{6.02} \\ = 0.5$

We have the final answer as

<h3>0.5 moles</h3>

Hope this helps you

6 0

2 years ago

Select the more electronegative element in each pair.a.Cl or Fb.Se or Oc.N or Asd.Na or Mg

liubo4ka [24]

Answer:

The answer to your question is below

Explanation:

Electronegativity is a measure of how strongly atoms attract electrons to themselves.

Process

Look for the electronegativity of each element and compare.

a) Cl = 3.16 F = 3.98 Fluorine has a higher electronegativity

b) Se = 2.55 O = 3.44 Oxygen has a higher electronegativity

c) N = 3.04 As = 2.18 Nitrogen has a higher electronegativity

d) Na = 0.93 Mg = 1.31 Magnesium has a higher electronegativity

6 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago