what problems occurs if the firework mixture is not pure(they are dirty or contaminated?​

Answer:

D

Explanation:

It's essentially just half and half, meaning the offspring would have the nose of the mother, the eyes of the fathers, basically crud like that

The statement that is true about the offspring of an organism Y that reproduces se*ually is that offspring would show some traits of the parent other than Y.

Se*ual reproduction is the type of reproduction that involves two organisms to produce an offspring.

The process of se*ual reproduction involves one male and one female organism of the same species.

This means that the offspring of organisms that reproduced se*ually will show traits of both organism involved.

Therefore, statement that is true about the offspring of an organism Y that reproduces se*ually is that offspring would show some traits of the parent other than Y.

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#SPJ2

Answer:

Hallo

Explanation:

I took that question

Answer:

z=16 and five p orbitals are filled and three s orbitals are filled

Answer:

Fast change

Explanation:

Weathering, erosion, and deposition are examples of slow change. Earthquakes, erupting volcanoes, and landslides happen fast. Therefore, they can change the Earth's surface very quickly.

Answer:

proton and neutron in region one and electron orbiting in region two is the answer

Answer:a

Explanation:

Answer:

Answervis A.

Explanation:

Because,so that other scientists can validate the results.

I hope it's helpful!

Answer:

1.

Solid: A, B, and C.

Liquid: D and E.

Gas: F.

2.

Into liquid on warm day (37 °C): D and C.

Into a liquid only when heated slightly: C and E.

From liquid to solid readily when placed into a freezer (0 °C): D.

Explanation:

Hello!

In this case, since the melting point is the temperature at which a substance change from solid to liquid and the boiling point from liquid to gas, we understand that substance at a temperature below the melting point is solid, whereas above are liquid; moreover, below the boiling point they are liquid and above gaseous.

In such a way, we have:

1.

Solid: A, B, and C.

Liquid: D and E.

Gas: F.

2.

Into liquid on warm day (37 °C): D and C.

Into a liquid only when heated slightly: C and E.

From liquid to solid readily when placed into a freezer (0 °C): D.

Best regards!

Answer:It creates a heat source on the cold water and the molecules act against each other and then it becomes steam.

Explanation:I had to take this on a test.

Answer:

C

Explanation:

Mol x Avogadro's number

Mol : mass / molar mass

Avogadro's no: 6.02E+23

In 15.0 g of sulfur, there are 2.82 × 10²³ atoms of S (C).

We want to calculate the number of atoms in 15.0 g of sulfur. We will need 2 conversion factors.

A conversion factor is an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

To convert 15.0 g of sulfur into atoms, we need to consider the following conversion factors:

[tex]15.0g \times \frac{1mol}{32.07g} \times \frac{6.02 \times 10^{23}atoms }{1mol} = 2.82 \times 10^{23} [/tex]

In 15.0 g of sulfur, there are 2.82 × 10²³ atoms of S (C).

Learn more about Avogadro's number here: https://brainly.com/question/15989275

Answer:

1. Mass of potassium (K) = 203.32 g

2. Number of mole of As = 7.53 moles

Explanation:

1. Determination of the mass of potassium (K)

Molar mass of K = 39.1 g/mol

Number of mole of K = 5.2 moles

Mass of K =.?

Mole = mass / Molar mass

5.2 = mass of K / 39.1

Cross multiply

Mass of K = 5.2 × 39.1

Mass of potassium (K) = 203.32 g

2. Determination of the number of mole of Arsenic (As)

Molar mass of As = 74.92 g/mol

Mass of As = 563.9 g

Number of mole of As =.?

Mole = mass /Molar mass

Number of mole of As = 563.9 / 74.92

Number of mole of As = 7.53 moles

Answer:

good luck with these problems I've never seen anything like it

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

[tex]\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100[/tex]    

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].          

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

[tex] Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]} [/tex]  (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

[tex] 1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x} [/tex]  

[tex] 1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0 [/tex]

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:      

[tex] \% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \% [/tex]                      

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

[tex] Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]} [/tex]

With [CH₃COOH] = 0.01 M

[tex] 1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x} [/tex]  

[tex]1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0[/tex]

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]    

Then, the percent ionization for this case is:

[tex] \% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \% [/tex]

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!    

The 0.010 M solution of CH₃COOH has a higher percent ionization compared to the 0.10 M solution.

The percent ionization of the acid solution is determined using the following equation:

Ionization equation of CH₃COOH is given below:

CH₃COOH + H₂O  ⇄ CH₃COO⁻ + H₃O⁺

Ka = 1.7 * 10⁻⁵

The  [H₃O⁺] in both solutions is determined first.

For the 0.1 M solution:

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

Using the equation of dissociation constant:

Ka =  [CH₃COO⁻][H₃O⁺]/[CH₃COOH]

1.7 * 10⁻⁵ = x₂/0.1 - x

assuming x is very small;

x = 1.29 * 10 ⁻³ M

Hence, [H₃O⁺] = 1.29 * 10 ⁻³ M

For the 0.010 M solution

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.01 - x                                         x                     x

Using the equation of dissociation constant:

Ka =  [CH₃COO⁻][H₃O⁺]/[CH₃COOH]

1.7 * 10⁻⁵ = x₂/0.01 - x

assuming x is very small;

x = 4.04 * 10 ⁻⁴ M

Hence, [H₃O⁺] = 4.04 * 10 ⁻⁴ M

For the 0.1 M solution

Percent ionization = (1.29 * 10 ⁻³ M/0.1) * 100%

Percent ionization = 1.29%

For the 0.01 M solution:

Percent ionization = (4.04 * 10 ⁻⁴ M/0.1) * 100%

Percent ionization = 4.04 %

Therefore, the 0.010 M solution of CH₃COOH has a higher percent ionization compared to the 0.10 M solution.

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Answer: aluminum oxide

Explanation:

Answer:

aluminum oxide

Explanation:

4.2 × 10⁻²¹ moles Sn

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

2500 atoms Sn

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

4.15144 × 10⁻²¹ moles Sn ≈ 4.2 × 10⁻²¹ moles Sn

Answer:

4

Explanation:

the actual change in phase is 4 on the diagram

Answer:

Explanation:

The lewis structure illustrates the bonding occuring between atoms of a molecule in conjunction with the lone pairs of electrons which have the possibility of existence in the molecule.

The image attached below shows the Lewis structure for the given Carbonyl dichloride (COCl₂) compound. From the image, the chlorine contains seven valence electrons, the carbon contains four valence electrons and oxygen contains six valence electrons.

Answer:

Pp, Pp, pp, pp, 50% chance at purple and 50% chance at white

Explanation:

Answer:

Group 18 (VIIIa)

Explanation:

Answer:

pure substance

Explanation:

it does not combine to anything.

Answer:

Cation

Anion

Cation

Cation

Explanation:

If an atom, or atoms, has a balanced number of electrons (negative charge) and protons (positive charge) they are neutral overall. However, if they are not balanced, they will be charged. These charged species are called ions.

What is an anion?

An anion has more electrons than protons, consequently giving it a net negative charge. For an anion to form, one or more electrons must be gained, typically pulled away from other atoms with a weaker affinity for them. The number of electrons gained, and so the charge of the ion, is indicated after the chemical symbol, e.g. chlorine (Cl) gains one electron to become Cl-, whilst oxygen (O) gains two electrons to become O2-.

What is a cation?

A cation has more protons than electrons, consequently giving it a net positive charge. For a cation to form, one or more electrons must be lost, typically pulled away by atoms with a stronger affinity for them. The number of electrons lost, and so the charge of the ion, is indicated after the chemical symbol, e.g. silver (Ag) loses one electron to become Ag+, whilst zinc (Zn) loses two electrons to become Zn2+.

Source: Cation vs Anion: Definition, Chart and the Periodic Table By  

Karen Steward (PhD)

Answer:

1. Solid

2. Melting Point

3. liquid

4. Boiling point

5. gas

Explanation:

Ionic bonds are special interatomic or intramolecular forces of attraction within a compound.

Answer:

Iron; [Ar] 3d6 4s2

Cobalt; [Ar] 3d6 4s2 4p1

Explanation:

We know that in our universe, there are five d orbitals known. However, in this strange universe L, there are only three d orbitals known.

The sixth and seventh first row transition elements are iron and cobalt. In universe L, the electronic configuration of iron and cobalt will be written as;

Iron; [Ar] 3d6 4s2

Cobalt; [Ar] 3d6 4s2 4p1

Answer:

Experiments should have the independent variable, and a dependent variable, as well as a constant variable and control variables.

Explanation:

Constants

Experimental constants are values that do not change either during or between experiments. Many natural forces and properties, such as the speed of light and the atomic weight of gold, are experimental constants. In some cases, a property can be considered constant for the purposes of an experiment even though it technically could change under certain circumstances. The boiling point of water changes with altitude and acceleration due to gravity decreases with distance from the earth, but for experiments in one location these can also be considered constants.

Independent Variable

The independent variable in an experiment is the variable whose value the scientist systematically changes in order to see what effect the changes have. A well-designed experiment has only one independent variable in order to maintain a fair test. If the experimenter were to change two or more variables, it would be harder to explain what caused the changes in the experimental results. For example, someone trying to find how quickly water boils could alter the volume of water or the heating temperature, but not both.

Dependent Variable

A dependent variable is what the experimenter observes to find the effect of systematically varying the independent variable. While an experiment may have multiple dependent variables, it is often wisest to focus the experiment on one dependent variable so that the relationship between it and the independent variable can be clearly isolated. For example, an experiment could examine how much sugar can dissolve in a set volume of water at various temperatures. The experimenter systematically alters temperature (independent variable) to see its effect on the quantity of dissolved sugar (dependent variable).

Control

A controlled variable is a variable that could change, but that the experimenter intentionally keeps constant in order to more clearly isolate the relationship between the independent variable and the dependent variable. For example, an experiment examining the relationship between how much sunlight plants receive (independent variable) and how tall they grow (dependent variable) should make sure none of the other factors change. The experimenter should control how much water the plants receive and when, what type of soil they are planted in, and as many other variables as possible.

The mole ratio  of ethanol to water : 1 : 3

Given

Reaction

C2H5OH + 302 →2C02 + 3H20

Required

The mole ratio

Solution

In a chemical reaction equation, the mole ratio of the reacting compound (reactants or products) is indicated by the reaction coefficient

The reaction coefficient is determined from the number of elements reacting by holding the principle that the number of elements in the reactants is equal to the number of elements in the product

In the above reaction, the ethanol-C2H5OH coefficient is 1 while the reaction coefficient of water-H20 is 3 so that the mole ratio = 1: 3

Answer:

Wow I don't that hhehehe sorry