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kirill115 [55]
3 years ago
15

What do all properties of matter have in common

Chemistry
2 answers:
VashaNatasha [74]3 years ago
8 0
No because some properties of matter are different from others
Wewaii [24]3 years ago
7 0

Answer:

they all matter

Explanation:

cause you matter :)

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Chemicals are classified as health hazards when they pose which of the following hazardous effects?
AlexFokin [52]

Answer:

poisoning, breathing problems, skin rashes, allergic reactions, allergic sensitisation, cancer, and other health problems from exposure.

Explanation:

many hazardous chemicals are also classified as dangerous goods.

4 0
2 years ago
What three tests must all theories pass to be considered a "proven" theory?
VladimirAG [237]

 In order to become a scientific theory the three categories that it must pass are the following: 


1) Can the phenomena be recreated in a laboratory setting? 


2) Can variables be changed, yet still result in like observations? 


3) Is the phenomena truly natural or was it the result of a man-made force enacting upon it?

7 0
3 years ago
What are some ways that groundwater is used in the United States currently?
FrozenT [24]
Ground is used for agriculture, sink water, hose water, and even drinking water (from the aqueducts).
7 0
3 years ago
Read 2 more answers
What is the halflife of a radioisotope if a 20-g sample becomes 10g after 16 hours
ohaa [14]

Answer:

T½ = 16hours

Explanation:

Final mass (N) = 10g

Initial mass (No) = 20g

Time (t) = 16hours

T½ = ?

T½ = In2 / λ

But λ = ?

In(N/No) = -λt

In(10/20) = -(λ * 16)

In(0.5) = -16λ

-0.693 = -16λ

λ = 0.693 / 16

λ = 0.0433

Note : λ is known as the disintegration constant

T½ = In2 / λ

T½ = 0.693 / 0.0433

T½ = 16hours

The half-life of the sample is 16hours

5 0
3 years ago
A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at
weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

-Q_1=Q_2

Mass of tin = m_1=18.5 g

Specific heat capacity of tin = c_1=0.21 J/g^oC

Initial temperature of the tin = T_1=97.38^oC

Final temperature = T_2=T

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2=75.7 g

Specific heat capacity of water= c_2=4.184 J/g^oC

Initial temperature of the water = T_3=21.52^oC

Final temperature of water = T_2=T

Q_2=m_2c_2\times (T-T_3)

-Q_1=Q_2

-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)

On substituting all values:

-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0
4 years ago
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