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3 years ago

Who proved that there are charged particles in an atom? dalton chadwick thomson bohr or rutherford?

Chemistry

1 answer:

Svetradugi [14.3K]3 years ago

5 0

Dalton Chadwick Thomson Bohr invented this theory.

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For my note plz help faster

stiv31 [10]

Where’s the question? Or

8 0

3 years ago

Se disuelven 10 g de acido carbónico H2CO3 en 150 g de agua. Determina la concentración % m/m de esa disolución?

sergey [27]

The question is: 10 g of carbonic acid H2CO3 are dissolved in 150 g of water. Determine the% m / m concentration of that solution?

Answer: The% m / m concentration of that solution is 6.66%.

Explanation:

Given: Mass of solute = 10 g

Mass of solvent = 150 g

Formula used to calculate the %m/m is as follows.

$Percent (m/m) = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Percent (m/m) = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{10 g}{150 g} \times 100\\= 6.66%$

Thus, we can conclude that the% m / m concentration of that solution is 6.66%.

6 0

3 years ago

How many moles of dinitrogen tetroxide (N2O4) are in 6.49 x 109 particles?

yan [13]

Answer: There are $1.08 \times 10^{-14}$ moles $N_{2}O_{4}$ present in $6.49 \times 10^9$ particles.

Explanation:

According to the mole concept, 1 mole of a substance contains $6.022 \times 10^{23}$ particles.

Hence, number of moles present in $6.49 \times 10^9$ particles are calculated as follows.

No. of moles = $\frac{6.49 \times 10^9}{6.022 \times 10^{23}}$ mol

= $1.08 \times 10^{-14}$ mol

Thus, there are $1.08 \times 10^{-14}$ moles $N_{2}O_{4}$ present in $6.49 \times 10^9$ particles.

6 0

3 years ago

Read 2 more answers

The pOH of a solution is 6.0. Which statement is correct?

Anastaziya [24]

If the pOH is 6.0, then the pH is 14.0 - 6.0 = 8.0 :)

5 0

3 years ago

If NaOH is added to 0.010 M Al3+, which will be the predominant species at equilibrium: Al(OH)3 or Al(OH)- 4 ? The pH of the sol

fredd [130]

Answer:

Al(OH)- 4,

Explanation:

NaOH added to 0.010 M Al3+

The predominant species at equilibrium will be = Al(OH)- 4, and this because sodium hydroxide ( NaOH ) is a base will readily form a stable complex ion with aluminum ion like ( Al( OH ) - 4 . also the higher the Kf value the more stable the complex ion becomes and the more soluble Al(OH)3 becomes

hence the predominant species at equilibrium is : Al(OH)- 4,

5 0

3 years ago