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Archy [21]
3 years ago
11

In order to change a solid to a liquid or a liquid to a gas, what is needed?

Chemistry
1 answer:
Roman55 [17]3 years ago
6 0
Answer: increase in thermal heat

explanation: In order to change the state, more energy is needed.
When you want to go from a lower state (solid) to a higher state (liquid) then you need more energy. and more energy means more heat in this case. if it was going from liquid to solid, you would need less thermal energy.
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Please help me with this. i also need to know what the word heat means.
aliina [53]
Heat is like the fire it’s hot.
I think it’s B
Sorry if I’m wrong
3 0
3 years ago
A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. Which of these compounds will precipitate out of this combined s
Archy [21]

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. BaS0₄ (s) will precipitate out of this combined solution.

Molecular equation

K2SO4(aq) + Ba(NO3)2(aq)  → BaSO4(s) + 2 KNO3(aq)?

This equation represents a double displacement (replacement) reaction, also called a metathesis reaction, in which the reactant ions exchange places to form new products. The general equation is:

A-B + C-D → A-D + C-B;

where A and C are cations, and B and D are anions.

Complete ionic equation: Includes all ions and the precipitate.

2K^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2[NO3]^-(aq) → 2K^+(aq) + 2[NO3]^- + BaSO4(s)

In an aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution.

The solid formed is called the precipitate. In case of an inorganic chemical reaction leading to precipitation, the chemical reagent causing the solid to form is called the precipitant.

Learn more about precipitation here : brainly.com/question/1783904

#SPJ4

6 0
1 year ago
Why does the rate of a reaction increase when the concentration of reactants is increased?
leva [86]

If reactants eventually collide, there is an occurrence of reaction.

<span> 
Therefore, when there is an increase concentration of reactant, meaning to say that there are several moles of it every unit volume. An example of this is a room having hundred of people will absolutely get higher concentration compared to a room with one individual only. 

Pertaining to effective collisions, if ever there is an increase of concentration, the frequency and rate of effective collisions among reactants surges in such a way that the rate of reaction also surges. Same with passing into a room with only 1 individual compared to hundred people blind persons, you probably want to proceed to the room with several people.</span>

<span>This is the simple logic behind that scientific existence.</span>

5 0
3 years ago
How many different elements are present in c6h12o6
Margarita [4]

Answer:

3 different elements are present in the compound.

5 0
2 years ago
Caffeine, a stimulant found in coffee and soda, hasthe mass percent composition: C. 49.48%, H, 5.19%. N. 28.85% 0. 16.48% The mo
borishaifa [10]

We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

C_4H_5O_1N_2

5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

C_8H_{10}O_2N_4

8 0
1 year ago
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