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3 years ago

Pls helllllpppppp!!!!! how was the interior of the atom was experimentally dicovered?

Chemistry

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons. Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup."

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Classify the elements as diatomic or monatomic. If necessary, refer to the periodic table to help you identify the diatomic elem

Oksanka [162]

Answer:

Carbon: Monatomic

Helium: Monatomic

Hydrogen: Diatomic

Iodine: Diatomic

Oxygen: Diatomic

Sodium: Monatomic

Explanation:

There you go

6 0

3 years ago

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How many moles of sulfur are needed to produce 15.0 mol of sulfur dioxide?

Brut [27]

<span>S + O2 = SO2

1 mole S ------------- 1 mole SO2
( moles S ) --------- 15.0 moles SO2

moles S = 15.0 x 1 / 1

moles S = 15.0 / 1

= 15.0 moles S</span>

4 0

3 years ago

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Why is the weather so hard to predict?

dusya [7]

Weather is very fast and often times unpredictable

3 0

3 years ago

3. During an experiment where 50.0 mL of a 1.0 M acid solution was mixed with 50.0 mL of a 1.0 M base solution, the temperature

Aleksandr-060686 [28]

Answer:

$\large \boxed{\text{-61 kJ$\cdot$mol$^{-1}$}}$

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50.0 50.0

c/mol·L⁻¹: 1.0 1.0

ΔT = 6.5 °C

ρ = 1.210 g/mL

C = 4.18 J·°C⁻¹g⁻¹

C_cal = 12.0 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}$

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

$\text{Mass of solution} = \text{100.0 mL} \times \dfrac{\text{1.10 g}}{\text{1 mL}} = \text{110.0 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0

0.0500ΔH + 2989 + 78.0 = 0

0.0500ΔH + 3067 = 0

0.0500ΔH = -3067

ΔH = -3067/0.0500

= -61 000 J/mol

= -61 kJ/mol

$\text{The enthalpy of reaction is $\large \boxed{\textbf{-61 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}$

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.

7 0

3 years ago

Can anyone help me on this please !

Julli [10]

Answer:

16.....................

3 0

3 years ago