A greater force is required to move an object with a larger mass than one with a smaller mass.

Chemistry

2 answers:

marishachu [46]3 years ago

7 0

Answer:

Explanation:

what do u need to know

charle [14.2K]3 years ago

5 0

The greater the force that is applied to an object of a given mass, the more the object will accelerate. The greater the mass of an object, the less it will accelerate when a given force is applied. For example, doubling the mass of an object results in only half as much acceleration for the same amount of force. Hope this helps!

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Please solve quickly!!!

vodka [1.7K]

The answer is c because it's the way there bonded together

3 0

3 years ago

[16]. A 3.0 L container holds a sample of hydrogen gas at 150 kPa. If the pressure increases to 2 atm and the temperature remain

boyakko [2]

First, let's convert atm to kPa:
2atm x 101.33kPa = 202.66kPa

Now, to find the pressure, you would have to use Boyle's Law: (P1V1)=(P2V2)
(150kPa x 3.0L) = (202.66kPa x V2)
V2= <span>(150kPa x 3.0L)/202.66kPa
V2=2.2L, B is your answer. This makes because volume and pressure have an inverse relationship, so since pressure increased volume should decrease.
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8 0

3 years ago

What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5

Step2247 [10]

CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x

$K=\frac{|CH_{3}COO^{-}|*|H^{+}|}{|CH_{3}COOH|}=\frac{x^{2}}{|CH_{3}COOH|}\\\\
1,78*10^{-5}=\frac{x^{2}}{0,06} \ \ |*0,06\\\\
0,1068*10^{-5}=x^{2}\\\\
x_{1}\approx0,001 \ \land \ \ x_{2}=\approx-0,001\\\\
pH=-log|H^{+}|=-log0,001=3$

4 0

3 years ago

Why are the elements placed in specific places on the periodic table?

harkovskaia [24]

•common trends and similarities
•similar ions going down (ex: Mg+2 and Ca+2)
•same number of electron orbitals going across

3 0

3 years ago

Read 2 more answers

Out of a box of 46 matches, 22 lit on the first strike. What percentage of the matches in the box did not light on the first str

ch4aika [34]

Answer: 52.17%

Explanation:

Number of matches in the box = 46

Number of matches that lit on the first strike = 22

Number of matches that did not light on the first strike = 46 - 22 = 24

Therefore, the percentage of the matches in the box did not light on the first strike will be:

= (Number of matches that did not light on the first strike / Number of matches in the box) × 100

= 24/46 × 100

= 52.17%

Therefore, the percentage of the matches in the box that did not light on the first strike is 52.17%.

7 0

3 years ago