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3 years ago

How does electromagnetism play a role in space

Chemistry

1 answer:

mr_godi [17]3 years ago

8 0

Answer:

Magnets can be used in space. ... One class of magnets, called electromagnets, does need electricity to work.

Explanation:

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LOTS OF POINTS (75) I WILL GIVE BRAINLIEST!!!!!

DedPeter [7]

Answer:

In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed. The distinction between physical and chemical change is not clear cut.

Explanation:

5 0

3 years ago

A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

Nitella [24]

Answer : The specific heat of metal is $0.481J/g^oC$ .

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

Now put all the given values in the above formula, we get

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

4 0

3 years ago

4. Rock is broken down as a result of chemical reactions;

tekilochka [14]

Answer:

Explanation:

CHEMICALS WEATHER IN CHEMICAL WEATHERING

7 0

3 years ago

Read 2 more answers

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

3 years ago

Calculate the density of a solid in g/cm3 if it weighs 38.3 kg and has a volume of 0.00463 m3.

RideAnS [48]

Answer: D=8.27 g/cm³

Explanation:

Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.

$D=\frac{38.3kg}{0.00463m^3} *\frac{(1m)^3}{(100cm)^3} *\frac{1000g}{1kg}$

With this equation, the m³ and kg cancel out, and we are left with g/cm³.

D=8.27 g/cm³

3 0

3 years ago