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insens350 [35]
3 years ago
15

Which of the following are likely to form a covalent bond?

Physics
2 answers:
sesenic [268]3 years ago
8 0
The answer is
"C"
<span>Magnesium and chlorine</span>
Vera_Pavlovna [14]3 years ago
4 0

its c- sulfur and oxygen trust me

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Its mass is 20 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume?
Alik [6]
Volume = mass / density
Volume = 20 / 7.87
Volume = 2.54 (2 s.f)
3 0
3 years ago
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
svetlana [45]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

6 0
2 years ago
How can groundwater flow, elevation can transport contaminates
Pavel [41]

Answer: it's under explanation

Explanation:Groundwater contamination occurs when man-made products such as gasoline, oil, road salts and chemicals get into the groundwater and cause it to become unsafe and unfit for human use. Materials from the land's surface can move through the soil and end up in the groundwater.

3 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
3 years ago
What property do all three of these common household substances have in common?
Oxana [17]

Answer:

whats the items

Explanation:

4 0
3 years ago
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