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3 years ago

One half of a balanced chemical equation is shown.

Physics

2 answers:

Zarrin [17]3 years ago

6 0

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

I took the test on Edgenuity

Tpy6a [65]3 years ago

6 0

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

i took the test responsibily

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28.25 mL, three signicant digits

Mazyrski [523]

Answer:3) variable affinities (stickiness) for something it is running past. Physical ... -measurement number (significant digits) unit (such as inches) -Significant ... Mass 1 oz. 28.25 g. Relations Between English and Metric Units Mass 1 dram. 1.772 g ... -graduated cylinder has an error of about 1% (± 0.1 mL in 10 mL). -Volumetric

Explanation:

3 0

3 years ago

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Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro

alex41 [277]

They were formed in the nuclear fusion reaction inside older stars.

As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.

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3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

Part B 

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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3 years ago

Which countries have high productivity, but are not among the top 10 for human development? Check all that apply.

trasher [3.6K]

Answer:

Belgium

France

Luxembourg

Explanation:

These are the ones that are in the High Productivity chart, but not in the HDI chart

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3 years ago

Help me please !!!!!

bezimeni [28]

Answer:

confounding cause they had exposure to many programmes

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3 years ago

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