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2 years ago

Which statement belongs to daltons atomic theory

1 answer:

5 0

Atoms cannot be broken down into smaller pieces is the statement belongs to daltons atomic theory. Option B is correct.

The atom consists of matter that may be split without releasing electrical charges. It's also the smallest unit of matter with chemical element features.

The complete question is:

"Which statements belong to Dalton's atomic theory? Select all that apply.

A)Atoms cannot be created or destroyed.

B)Atoms cannot be broken down into smaller pieces.

C)All atoms of an element are identical.

D)Chemical reactions cause atoms to be rearranged."

According to Dalton's atomic hypothesis, atoms cannot be broken down into smaller bits.

Hence, option B is correct.

To learn more about the atom refer to the link;

brainly.com/question/1566330

#SPJ1

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The period T of a simple pendulum depends on the length L of the pendulum and the acceleration of gravity g (dimensions L/P). (a

andriy [413]

Answer:

a). L= meters, g= $\frac{m}{s^{2} }$

b). L= 5cm T=0.448

L= 10 m T=6.34

c). Constant= $\frac{2*\pi }{\sqrt{g} }=\frac{2*\pi }{\sqrt{9.8} }=2.007089923$

Explanation:

a).

$T= 2*\pi \sqrt{\frac{L}{g} } = 2*\pi \sqrt{\frac{m}{\frac{m}{s^{2} } } } \\T= 2*\pi \sqrt{\frac{s^{2}*m }{m} }=2*\pi \sqrt{s^{2} } \\T= s$

b).

$L_{1}= 5 cm$ , $5cm *\frac{1m}{100 cm} = 0.05 m$

$T=2*\pi \sqrt{\frac{L}{g} }$

$T=2*\pi \sqrt{\frac{0.05}{9.8} }= 0.448$ s

$L_{1}= 10 m$

$T=2*\pi \sqrt{\frac{L}{g} }$

$T=2*\pi \sqrt{\frac{10}{9.8} }= 6.43$ s

c).

$g= 9.8 \frac{m}{s^{2} }$

$T=2*\pi *\frac{\sqrt{L} }{\sqrt{g} } =T=2*\pi *\frac{\sqrt{L} }{\sqrt{9.8} } \\T= 2*\pi \frac{1}{\sqrt{9.8}} *\sqrt{L}\\T= 2.007089923*\sqrt{L}$

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3 years ago

Which element causes a star to go through a high mass cycle

denis23 [38]

Answer: As larger stars reach the red-giant phase, their temperature increases as helium is fused into carbon. Core temperature increases, with fusion forming oxygen, nitrogen and iron. When the star core converts to iron, fusion ceases.

Explanation:

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3 years ago

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3m3 and the pressure increases from 2.50×105

SVETLANKA909090 [29]

Answer:

$W = -4.95 \times 10^4\ J$

Explanation:

given,

In first case Volume remains constant.

Work done in the first case is zero.

In Second case Volume change

V₁ = 0.2 m³

V₂ = 0.11 m³

Pressure, P = 5.5 x 10⁵ Pa

Work done = Pressure x change in volume

W = P ΔV

$W = P(V_2-V_1)$

$W = 5.5\times10^5\times (0.11 - 0.2)$

$W = -4.95 \times 10^4\ J$

Hence, Work done when volume changes is equal to $W = -4.95 \times 10^4\ J$

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3 years ago

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3 years ago

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