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guapka [62]
3 years ago
7

Help me please, I have no clue but hey first day of school

Chemistry
1 answer:
sineoko [7]3 years ago
3 0

kilo is 1,000 times the base unit <em>(tip for remembering: think money, 1k is 1,000)</em>

centi is 1/100 of the base unit <em>(tip for remembering: cent = 100)</em>

micro is 1/1,000,000 of the base unit

nano is 1/1,000,000,000 of the base unit <em>(tip for remembering: </em><em>n</em><em>ano will have </em><em>n</em><em>ine zeroes)</em>

milli is 1/1,000 of the base unit

mega is 1,000,000 times the base unit


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Identify the compounds that are soluble in both water and hexane.
dlinn [17]

Answer:

The compound which are boh soluble in water and hexane is

B. Ethanol and 1-propanol  

Explanation:

The compounds ethanol and 1-propanol are soluble in both hexane and water.

It is soluble in water as both  consists of polar end due to hydrogen bonding present in the  -OH functional group.

and both are soluble in hexane as they contain a non polar end and the alliphatic hydrocarbon chain in them.

The solubility of alcohols varies in increasing order as the hydrocarbon chain increases. And becaue of this it becomes more non polar.

Non polar properties decreases for branched molecules.

so, the correct option is ethanol and 1-propanol.

3 0
3 years ago
The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes su
masha68 [24]

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ T(^{\circ} C)=T(K)-273.15

By putting the values, we get

⇒            =4.483-273.15

⇒            =-268.667^{\circ} C

Thus the above is the correct answer.

3 0
3 years ago
Which of the following would not be an isotope of carbon-12? A. Carbon-14 B. Carbon-13 C. Boron-12 D. Carbon-11
Oliga [24]
C because it isn't carbon
6 0
3 years ago
Describe how you would set up an experiment to test the rela-tionship between completion of assigned homework and the fi-nal gra
ValentinkaMS [17]

To start this test, you need to identify the variables it presents. As you may already know, there are independent and dependent variables. Independent variables are those that act on a factor, influencing it to generate a result. In the case of this experiment, the independent variable is the completion of the homework. The dependent variable, in turn, is the factor that receives the influence of the independent variable, in this experiment this variable is the final grade you received in the course.

After that you must select a number of students, give them their homework and ask each student to complete a percentage of that amount. An example of this could be that you select 11 students and ask the first to complete 0% of the homework, the second student must complete 10%, the third 20% and so on, and the 11th student must complete 100% of the homework.

after that, note what was the final grade that each student received in the course and make a graph to show the results.

The y-axis of the graph must represent the dependent variable, while the x-axis must represent the independent variable. This way you will show the exact relationship between completing homework and the final grade of the course.

7 0
3 years ago
How many moles in 4.93 x 10E23 atoms of silver?
leva [86]
<h3>Answer:</h3>

0.819 mol Ag

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})
  2. Divide:                              \displaystyle 0.818665 \ mol \ Ag

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.818665 mol Ag ≈ 0.819 mol Ag

8 0
2 years ago
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