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3 years ago

Which of the following is a heterogeneous mixture? A. salt B. dye in water C. sugar water D. a garden salad

Physics

1 answer:

dedylja [7]3 years ago

7 0

D garden salad : )

A heterogenous mixture can be easily taken apart visually/physically

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The number of electrons an atom would gain or lose when forming ionic bonds. It can be positive, negative or zero.

vaieri [72.5K]

Answer:

In an ionic bond , an element will have to lose or gain electrons.

Explanation:

Ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.
Such a bond forms when the valence (outermost) electrons of one atom are transferred permanently to another atom.
The atom that loses the electrons becomes a positively charged ion (cation), while the one that gains them becomes a negatively charged ion (anion).

∴

The number of electrons an atom would gain or lose when forming ionic bonds cannot be zero.

8 0

3 years ago

A country would place a tariff on imported steel to

Whitepunk [10]

. . . 'protect' its domestic steel industry, by
increasing the price of imported steel.

7 0

3 years ago

Read 2 more answers

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

3 years ago

Hàng ngày ta thấy Mặt trời , Mặt trăng quay quanh Trái Đất, khi đó ta đã lấy vật mốc là

marshall27 [118]

I’m sorry I don’t understand this language

5 0

3 years ago

During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun

Ray Of Light [21]

Answer:

At the closest point

Explanation:

We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"

In this problem, we have a comet orbiting around the Sun:

- Its closest distance from the Sun is 0.6 AU

- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: $A\propto vr$ , therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster ( $A\propto vr$ , if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.

7 0

3 years ago