1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
noname [10]
3 years ago
15

____ ha no nucleus.

Chemistry
2 answers:
Serga [27]3 years ago
7 0

I think its prokaryotes?

SVETLANKA909090 [29]3 years ago
5 0
Prokaryotes...hope this helped
You might be interested in
The molar concentration (M) of a solution prepared by dissolving 0.2362g of Cr(NO3)3 in a 50-mL volumetric flask is 0.01985M, wh
Vinil7 [7]

Answer:

Here's what I get  

Explanation:

You want to dilute the original solution by a factor of 25 in two steps, so you could dilute it by a factor of 5 in the first step, then dilute the new solution by another factor of 5.

A. First dilution

Use a 10 mL pipet to transfer 10 mL of the original solution to a 50 mL volumetric flask. Make up to the mark with distilled water. Shake well to mix.

Use the dilution formula to calculate the new concentration.

\begin{array}{rcl}c_{1}V_{1} & = & c_{2}V_{2}\\0.01985 \times 10.00 & = & c_{2} \times 50.00\\0.1985 & = & 50.00 c_{2}\\\\c_{2}& = & \dfrac{0.1985}{50.00}\\\\& = & \text{0.003 970 mol/L}\\\end{array}

B. Second dilution

Repeat Step 1, using the 0.003 970 mol·L⁻¹ solution.

\begin{array}{rcl}c_{2}V_{2} & = & c_{3}V_{3}\\0.003970 \times 10.00 & = & c_{3} \times 50.00\\0.03970 & = & 50.00 c_{3}\\\\c_{3}& = & \dfrac{0.03970}{50.00}\\\\& = & \textbf{0.000 7940 mol/L}\\\end{array}\\\text{The concentration of the final solution is $\boxed{\textbf{0.000 7940 mol/L}}$}

3. Check:

Compare the final concentration with the original

\begin{array}{rcl}\dfrac{ c_{3}}{ c_{1}} & = & \dfrac{0.0007940}{0.01985}\\& = & \mathbf{\dfrac{1}{25.00}}\\\end{array}\\\text{The concentration of the final solution is } \boxed{\mathbf{\dfrac{1}{25}}} \text{ that of the original solution}

7 0
3 years ago
What is the mass of 6.02 x 1023 particles of rubidium carbonate
Firdavs [7]

Answer:

Explanation:

84.97

4 0
3 years ago
8. How does the arctic fox survive the harsh winters in the tundra?
Ket [755]

Answer:

the answer is d

Explanation:

It migrates to coastal areas during the winter months to catch fish.

4 0
3 years ago
Please go to this link... I've been waiting for someone to answer for 4 hours...
Charra [1.4K]
On my way! to find the answer
6 0
4 years ago
What the common uses for Bohrium?
lora16 [44]

Answer:

Explanation:

Bohrium's most stable isotope, bohrium-270, has a half-life of about 1 minute. It decays into dubnium-266 through alpha decay. Since only a few atoms of bohrium have ever been made, there are currently no uses for bohrium outside of basic scientific research.

5 0
3 years ago
Read 2 more answers
Other questions:
  • A 74.0-gram piece of metal at 94.0 °C is placed in 120.0 g of water in a calorimeter at 26.5 °C. The final temperature in the ca
    12·1 answer
  • A solution is prepared by adding 0.700 g of solid NaClNaCl to 50.0 mL of 0.100 M CaCl2CaCl2. What is the molarity of chloride io
    12·1 answer
  • What does sand grab onto to help form beaches?
    7·2 answers
  • During which process of the water cycle does water change from a gas to a liquid? A.precipitation
    12·2 answers
  • Decide if the following statements regarding Intermolecular forces are True or False. H2S will have a higher boiling point than
    14·1 answer
  • Is comets greater than asteroids
    11·2 answers
  • The opposite of vaporization is called
    12·2 answers
  • Sulfur and fluorine form several different compounds including sulfur hexafluoride and sulfur tetrafluoride. Decomposition of a
    15·1 answer
  • If you wished to displace cobalt(II) ions from
    8·1 answer
  • The actual density of iron is 7.874 g/mL. In a laboratory investigation, Jason finds the density of a piece of iron to be 7.921
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!