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2 years ago

What would cause a liquid to turn into a solid?

Chemistry

2 answers:

alexira [117]2 years ago

7 0

Answer:

Freezing the liquid

Explanation:

Freezing a liquid causes a liquid to become a soild.

cricket20 [7]2 years ago

3 0

Answer:

changing the temperature of the liquid

Explanation:

Ex. Cooking brownie batter(liquid), turns it into an edible brownie(solid). Ex. Putting water into a freezer freezes it, turning it into a solid.

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What is the activation energy of this reaction? 10 kJ 25 kJ 30 kJ 35 kJ

Nostrana [21]

Answer:

25 kj

Explanation:

8 0

3 years ago

How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106

il63 [147K]

Hey there :)

Question: How many km are in 5.6mm?

=>5.6x10^3

=>5.6x10^-6

=>5.6x10^-3

=> 5.6x10^6

Answer:-

$5.6 \times 10^{ - 6}$

Explanation:-

By using the formula-

$1millimeter = \frac{1}{1000000}$

As 1 with 6 zeros, we convert it into exponential form.

$= > \frac{1}{10^{6} }$

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

$= > 10^{ - 6}$

Now combine with given question.

$= > 5.6 \times 10^{ - 6}$

7 0

2 years ago

Which two celestial bodies orbit the Sun?

tiny-mole [99]

Number 1 is Comets and asteroids
Number 2 is Comets are made of ice and rock
Number 3 is Meteor

7 0

3 years ago

Read 2 more answers

What are the steps to finding an items density?

Assoli18 [71]

There are 5 steps:
1- measure the mass of the container
2- measure the volume of the liquid
3- measure the combined mass of the liquid and the container.
4-determine the mass of the liquid alone
5-divide the mass by the volume.

3 0

2 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