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MakcuM [25]
2 years ago
15

Single atoms can be seen with a powerful light microscope

Chemistry
2 answers:
AysviL [449]2 years ago
8 0

Answer:

The answer is false

Explanation:

marysya [2.9K]2 years ago
5 0
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3. What is the name of the technique used to weigh the unknown liquid in part A of the
Anna [14]

Answer:

The technique is called weighing by difference

7 0
3 years ago
Who preformed the oil drop experiment
Ivan

Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.

5 0
2 years ago
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17. Which of the following is NOT an empirical formula? *
Aloiza [94]

Answer: 17) d. C_2H_6

18. c. The empirical formula of a compound can be twice the molecular formula.

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

The empirical mass can be calculated from empirical formula and molar mass must be known.

17. Thus the empirical formula of C_2H_6 should be CH_3

18. The molecular formula will either be same as empirical formula or is a whole number multiple of empirical formula. Thus the empirical formula of a compound can never be twice the molecular formula.

3 0
3 years ago
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What is the pH of a Koh solution that has [H+]=1.87×10^-13M
xenn [34]
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72

I hope this helps.  Let me know if anything is unclear.
7 0
3 years ago
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A chemist prepares a solution of sodium carbonate by measuring out of sodium carbonate into a volumetric flask and filling the f
Vadim26 [7]

Answer:

The correct answer will be "9.45\times 10^{-5} \ mol/L".

Explanation:

We have:

Moles of solute (Na₂Co₃) = 18.9 \mu \ mol

                                           = 18.9\times 10^{-6}

                               1 \mu \ mol = 10^{-6} \ mol

Now,

The volume of the solution will be:

⇒  0.200 \ L

∴ {1 \ mL=0.001 \ L}

The formula to find the concentration will be:

⇒  Concentration = \frac{moles \ of \ solute \ (mol)}{Volume \ of \ solution \ (L)}

On substituting the given values, we get

⇒                          =\frac{18.9\times 10^{-6}}{0.200}

⇒                          =9.45\times 10^{-5} \ mol/L

8 0
3 years ago
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