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Tom [10]
4 years ago
13

Convert 59,800 cg/L to g/mL

Chemistry
1 answer:
Shkiper50 [21]4 years ago
3 0
Answer:
0.598 g/ml

Explanation:
From the basics of conversions:
1 cg is equivalent to 0.01 g
1 liter is equivalent to 1000 ml
Therefore:
To convert cg to g, you have to multiply  by 0.01
To convert L to ml, you have to multiply by 1000
Therefore:
cg/L to g/ml = 0.01/1000 = 1*10^-5
This means that:
To convert from cg/l to g/ml, you have to multiply by 1*10^-5

Using the given:
59,800 cg/l = 59,800 * 1 * 10^-5 = 0.598 g/ml

Hope this helps :)

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A closed system is completely closed to the outside environment. Every interaction is transmitted inside that closed system.

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Which of the following is true for balancing equations? A. The number of products should be equal to the number of reactants. B.
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D.) There must be an equal number of atoms of each element on both sides of the equation

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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>

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<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})
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<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

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3 years ago
How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
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Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

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