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mihalych1998 [28]
2 years ago
13

What is space biology?? ​

Chemistry
1 answer:
Alex2 years ago
4 0

Answer:

space biology is a fundamental component of space life science. it focuses on smaller organisms such as cell, animals and plants.

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Identify each of the following compounds as: (a) a nonelectrolyte: lactose (c12h22o11) potassium chloride (kcl) lactic acid (hc3
11Alexandr11 [23.1K]

Nonelectrolytes are compounds that do not tend to get ionize in all solution. As a result, solutions which contains nonelectrolytes will not conduct electricity.

Ionic compounds dissolve in water and break into the constituent ions through the process called ionization. Now These electrolyte solutions can either be acids, or bases, or salts.

If  the process of ionization produces a large no. of ions in the solution, it is a strong electrolyte. On the other hand, if it produces only a few ions, the solution is a weak electrolyte,

Potassium chloride (KCL) is a strong electrolyte,

Lactic acid (HC3H503) is a weak electrolyte,

Whereas, Ethanol (C3H7OH) is a non electrolyte,

Dimethylamine (CH3)2NH is a weak electrolyte,

Sodium hydroxide(NaOH) is a strong electrolyte.

To learn more about nonelectrolytes, head here

brainly.com/question/3712208

#SPJ4

6 0
1 year ago
8.03 Solutions Lab Report<br> Does anyone have a PDF or Document of FLVS 8.03 Solutions Lab
GarryVolchara [31]

8.03 solutions report is described below.

Explanation:

8.03 Solutions Lab Report

In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.  

Pre-lab Questions:

In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.  

Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).

Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.

7 0
3 years ago
Read 2 more answers
Helium gas in a cylinder is under 1.12atm pressure at 25.0C. What will be the pressure if the temperature increases to 37.0C?
Irina18 [472]

Answer:

p_2=1.17atm

Explanation:

Hello!

In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:

\frac{p_2}{T_2} =\frac{p_1}{T_1} \\\\

Thus, for the initial pressure and temperature in kelvins the final temperature in kelvins, we compute the final pressure as:

p_2=\frac{p_1T_2}{T_1} \\\\p_2=\frac{1.12atm*310.15K}{298.15K}\\\\p_2=1.17atm

Best regards!

7 0
2 years ago
5. What happens to the average kinetic energy of water molecules as water freezeslt decreases as the water absorbs energy from i
4vir4ik [10]
The correct answer among the choices listed above is option D. The average kinetic energy of water molecules as water freeze <span>decreases as water releases energy to its surroundings. Energy is released as the molecules go into a more condensed phase which is the solid.</span>
3 0
3 years ago
Read 2 more answers
g Given that 50.0 mL of 0.100 M magnesium bromide reacts with 13.9 mL of silver nitrate solution according to the unbalanced equ
ipn [44]

Answer:

0.719M AgNO₃

Explanation:

Based on the reaction:

MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂

<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>

<em />

To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:

<em>Moles AgNO₃:</em>

<em />

Moles of MgBr₂ are:

50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.

As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:

0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =

0.0100 moles of AgNO₃ are in the solution.

And molarity is:

0.0100 moles AgNO₃ / 0.0139L =

<h3>0.719M AgNO₃</h3>
3 0
3 years ago
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