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Leni [432]
3 years ago
12

Calculate the mean of 2, 5 and 3. write your answer to 2 decimal places. please help me on this

Chemistry
1 answer:
makvit [3.9K]3 years ago
6 0

Hello.

The answer is: 3.3333

To get the answer add 2,5 and 3 that is 10 then divide by 3 to get 3.3

Have a nice day

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Due tonight! Please help!
olchik [2.2K]

Answer:

42.98 g/L

Explanation:

6 0
3 years ago
What is the mass of 3.77mol of K3N?
AleksAgata [21]
<h3>Answer:</h3>

495 g K₃N

<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>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

3.77 mol K₃N

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

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

  1. Set up:                       \displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})
  2. Multiply/Divide:         \displaystyle 495.039 \ g \ K_3N

<u>Step 4: Check</u>

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

495.039 g K₃N ≈ 495 g K₃N

5 0
3 years ago
When it acts as a base, which atom in hydroxylamine accepts a proton?
Rashid [163]
<span>The oxygen atom accepts the proton. The oxidation number of O is -2, meaning that there are two unshared electrons in the valence shell; In the ClO- ion, one of these is shared with the Cl- ion, leaving an unshared electron on the oxygen atom, which is what the hydrogen atom shares its electron with, becoming the proton accepted by the O atom.</span>
6 0
3 years ago
fish tank initially contains 35 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 5 l
weeeeeb [17]

Answer:

Therefore, the rate of change in the amount of salt is \frac{dx}{dt} =( 5c}{ - \frac{x }{20})

\frac{grams }{min}

Explanation:

Given:

Initial volume of water V = 35 lit

Flowing rate = 5 \frac{Lit}{min}

The rate of change in the amount of salt is given by,

   \frac{dx}{dt} = ( Rate of salt enters tank - rate of sat leaves tank )

Since tank is initially filled with water so we write that,

x(0) = 0

Let amount of salt in the solution is c,

  \frac{dx}{dt} = \frac{5c}{1 } - \frac{x(t) \times 5}{100}

  \frac{dx}{dt} =( 5c}{ - \frac{x }{20}) \frac{grams}{min}

Therefore, the rate of change in the amount of salt is \frac{dx}{dt} =( 5c}{ - \frac{x }{20})

\frac{grams }{min}

7 0
3 years ago
how are the relationships of elements in a group different from the relationships of elements in a group
alex41 [277]
Same group = same properties, same # of valence electrons
Same period = same # of atomic orbitals
3 0
3 years ago
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