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Zanzabum
3 years ago
11

What is the answers to these questions?

Chemistry
1 answer:
serious [3.7K]3 years ago
4 0

2. B

3. B

4. A

5. C

6. B

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A bowling ball (mass = 7.2 kg, radius = 0.10 m) and a billiard
evablogger [386]

Answer:

Maximum gravitational Force: F_{gmax} = 1,026*10^{-08} N

Explanation:

The maximum gravitational force is achieved when the center of gravity are the closer they can be. For the spheres the center of gravity is at the center of it, so the closer this two centers of gravity can be is:

bowling ball radius + billiard ball radius = 0,128 m

The general equation for the magnitude of gravitational force is:

F_{gmax}  = G \frac{M*m}{r_{min}^{2} }

Solving for:

G = 6,67*10^{-11}  \frac{Nm^{2}}{kg^{2}}

M = 7,2 kg

m = 0,35 kg

r_{min} = 0,128 m

The result is:

F_{gmax} = 1,026*10^{-08} N

3 0
3 years ago
A mixture that results when substances dissolve to form a homogeneous mixture
MrRa [10]
A mixture that results when substances dissolve to form a homogeneous mixture is a solution.
3 0
3 years ago
15
Vinil7 [7]

Answer:

barium and silicon has same valence electrons

Explanation:

barium-2,8,18,18,8,2

neon-2,8

silicon-2,8,2,2

carbon-2,4

5 0
2 years ago
What does volcanic ash add to the environment?
Travka [436]
Somethings that are beneficial is its good for plants, if it is very fine ash it is able to break down and quickly dissolve into the soil.
7 0
3 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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