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stiks02 [169]
3 years ago
11

HURRY PLEASE (MULTI CHOICE)

Chemistry
2 answers:
Mama L [17]3 years ago
6 0
The magnets in a maglev train system help move, or propel, the train
Lera25 [3.4K]3 years ago
5 0

Answer:

maglev

Explanation:

<em>In</em><em> </em><em>a</em><em> </em><em>train</em><em> </em><em>system</em><em> </em><em>two</em><em> </em><em>magnets</em><em> </em><em>are</em><em> </em><em>present</em><em>:</em><em>one</em><em> </em><em>set</em><em> </em><em>to</em><em> </em><em>repel</em><em> </em><em>and</em><em> </em><em>push</em><em> </em><em>the</em><em> </em><em>train</em><em> </em><em>up</em><em> </em><em>off</em><em> </em><em>the</em><em> </em><em>track</em><em>,</em><em>and</em><em> </em><em>another</em><em> </em><em>set</em><em> </em><em>the</em><em> </em><em>elevated</em><em> </em><em>train</em><em> </em><em>ahead</em><em>,</em><em>taking</em><em> </em><em>advantage</em><em> </em><em>of</em><em> </em><em> </em><em>the</em><em> </em><em>lack</em><em> </em><em>of</em><em> </em><em>friction</em><em>.</em>

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How does burning affect an object's property? Burning can cause an object to increase in volume. Burning changes the chemical ma
Jobisdone [24]

Answer: Burning changes the chemical make up of an object.

Explanation:

A chemical change can be defined as a change in the substance when it combines with other kind of substance to form a new substance. A chemical change can also occur when a substance is broken down into two or more products. These changes cannot be reversed. These changes affect the physical make up of an object. For example, burning as when an object is burned it cannot be transformed into its original form. A wood if burned can be converted into ash, water and carbon dioxide cannot regain its original form after burning so burning brings about chemical change in an object.

3 0
3 years ago
What are two factors that affect an objects kinetic energy
Bad White [126]
Potential energy and height; best guess;)
8 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
2 years ago
What do you think the density of a person might be? Explain
Verizon [17]
So this can explain the density of a person very clearly.

7 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li
meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ

4 0
2 years ago
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