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guapka [62]
3 years ago
6

The strength of the gravitational force between two objects depends on __________.

Chemistry
2 answers:
irga5000 [103]3 years ago
6 0
Should be their masses. Because t<span>he strength of the gravitational force between two objects depends on two factors, mass and </span>distance<span>. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.</span> 
RUDIKE [14]3 years ago
4 0
D. The larger the object the greater the force. That's why we are pulled toward the ground because Earth is so large compared to us.

It may not seem like it, but we are pulling the Earth, too. But the force is so much smaller, that the net value pulls us toward Earth more so.
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Describe how the 20 amino acids differ from each other.
sleet_krkn [62]

The difference between all of them in the make-up of the R group.



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6 0
3 years ago
Read 2 more answers
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
CH3-CH2-CH=CH-CH=CH2
o-na [289]

Answer:

Hydration (of an alkene)

Mechanism : Electrophilic addition.

8 0
3 years ago
All the examples stated belos is an example of electric forces except _________. A)a neutron pushing on another neutron B)an ele
Mandarinka [93]
All of the following are examples of electric force except A. A Neuton pushing on another Neutron. Neutrons are subatomic particles that possess no electrical charge.
4 0
3 years ago
Read 2 more answers
How many significant figures are in each of the following measurements? (c) 0.03003 kg (f) 3.8200 x 103 L (a) (d) 0.00450 m 35.0
erica [24]

Answer:

c =four significant figures = 3003

f = Four significant figures = 3820

d =(i) three significant figures = 450

d =(ii) Six significant figures  = 35.0445

b = significant figures = 59.0001

e = five significant figures = 67000

Explanation:

c = 0.03003 kg

four significant figures = 3003

f = 3.8200 × 10³ L

Four significant figures = 3820

d =(i) 0.00450 m and (ii) 35.0445 g

(i) three significant figures = 450

(ii) Six significant figures  = 35.0445

b = 59.0001 cm

six significant figures = 59.0001

e = 67000

five significant figures = 67000

The given measurement have five significant figures 41006. this measurement can be rounded to three significant figures i.e, 0.410 g.

All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.

Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.

Zero between the non zero digits are consider significant like 104 consist of three significant figures.

The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.

8 0
3 years ago
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