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horsena [70]
3 years ago
13

3.04 Quiz: Chemical Reactions 1 to 10 question

Chemistry
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer:

1. Chemical reactions

2.Substances

3. Properties

4. Precipitate

5. Light

6. Temperature

7. Color

8. Gas

Explanation:

During chemical reactions, new substances that possess new properties are formed. During the process of chemical reactions, evidences are found. Some of the evidences are formation of precipitate or light gel, production of gases and changes in color and temperature.

Chemical reactions start with the substances or compounds which are known as reactants. The reactants react to form new substances known as the products which possess new properties.

I only could help with 1 - 8 sorry.

You might be interested in
Position (m)
Alex777 [14]

Option C. The object is returning to the start at a constant speed.

<h3>Data points of the Position vs Time graph</h3>

The following data points will be used to determine the motion of the object.

<u>Position               Time</u>

12                          4

10                          6

2                            8

0                           10

From the data above, the position of the object is decreasing towards zero or start point.

Thus, the object is returning to the start at a constant speed.

Learn more about position here: brainly.com/question/2364404

#SPJ1

8 0
2 years ago
28. Which statement is INCORRECT regarding radiation?
mash [69]

Answer:

Radiation is energy. It can come from unstable atoms that undergo radioactive decay, or it can be produced by machines. Radiation travels from its source in the form of energy waves or energized particles. There are different forms of radiation and they have different properties and effects.

Related information in Spanish (Información relacionada en español)

On this page:

Ionizing and non-ionizing radiation

Electromagnetic spectrum

Types of ionizing radiation

Periodic Table

Non-Ionizing and Ionizing Radiation

There are two kinds of radiation: non-ionizing radiation and ionizing radiation.

Non-ionizing radiation has enough energy to move atoms in a molecule around or cause them to vibrate, but not enough to remove electrons from atoms. Examples of this kind of radiation are radio waves, visible light and microwaves.

Ionizing radiation has so much energy it can knock electrons out of atoms, a process known as ionization. Ionizing radiation can affect the atoms in living things, so it poses a health risk by damaging tissue and DNA in genes. Ionizing radiation comes from x-ray machines, cosmic particles from outer space and radioactive elements. Radioactive elements emit ionizing radiation as their atoms undergo radioactive decay.

Radioactive decay is the emission of energy in the form of ionizing radiationHelpionizing radiationRadiation with so much energy it can knock electrons out of atoms. Ionizing radiation can affect the atoms in living things, so it poses a health risk by damaging tissue and DNA in genes.. The ionizing radiation that is emitted can include alpha particles, beta particles and/or gamma raysHelpgamma raysA form of ionizing radiation that is made up of weightless packets of energy called photons. Gamma rays can pass completely through the human body; as they pass through, they can cause damage to tissue and DNA.. Radioactive decay occurs in unstable atoms called radionuclides.

Explanation:

8 0
3 years ago
The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.
adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol

Now we have to calculate the energy required.

Q=\frac{\Delta H}{n}

where,

Q = energy required

\Delta H = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0
3 years ago
Brainliest for correct answer please show all work
Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

<u>Question 1:</u>

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

<em>Balance the number of Na:</em>

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

<em>Balance the number of K:</em>

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

<em>The equation is now balanced.</em>

<u>Question 2:</u>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

<em>Balance</em><em> </em><em>n</em><em>u</em><em>m</em><em>b</em><em>e</em><em>r</em><em> </em><em>of</em><em> </em><em>L</em><em>i</em><em> </em><em>and</em><em> </em><em>F</em><em> </em><em>atoms</em><em>:</em>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

<u>Question 3:</u>

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>P</em><em> </em><em>atoms</em><em>:</em>

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>O</em><em> </em><em>atoms</em><em>:</em>

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

4 0
3 years ago
lucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of approximately 0.200 mM. i) What is
Mkey [24]

Answer:

The concentration is 0.036 mg/mL

Explanation:

Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL

3 0
3 years ago
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