is the prime meridian a longitudinal or latitudinal line?

Answer:

pure substance

Explanation:

it does not combine to anything.

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).

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Answer:

 59.4 tons of iron can be isolated from this ore

Explanation:

mass of ore = 250 tons

mass of hematite = 34 % of 250

= 85 tons

mass of iron = 69.9 % of hematite

= 69.9 % of 85

= 59.4 tons

So,  59.4 tons of iron can be isolated from this ore .

Answer:

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

Explanation:

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:

Part A:

First, convert molarity to moles by multiplying by the volume:

0.293 M AgNO3 = (0.293 moles AgNO3)/1 L x 1.19 L = 0.349 moles AgNO3

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:

V = 3.1 L      

Explanation:

Given data:

Molarity of solution = 0.37 M

Mass of LiF = 29.53 g

Volume of solution = ?

Solution:

Number of moles of LiF:

Number of moles = mass/molar mass

Number of moles = 29.53 g/ 25.94g/mol

Number of moles = 1.14 mol

Volume:

Molarity = number of moles of solute / Volume in L

0.37 M = 1.14 mol / V

V = 1.14 mol / 0.37 M

V = 3.1 L         (M = mol/L)

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:

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

Explanation:

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

Answer:

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

Answer:

hydrogen and oxygen

Explanation:

oxygen gas ignites a glowing splint because it supports burning while hydrogen gas extinguishes the burning splint with a pop sound.

the mixture of the two explodes thus the observation

Answer: Their main aim was to provide goods, specifically furs and sugar exportation.

Explanation:

Answer:

5% is the percent error

Explanation:

Percent error is defined as the 100 times the ratio of absolute value of the difference between the measurement and the theoretical value and the theoretical value, that is:

[tex]\frac{|Theoretical-Measurement|}{Theoretical} * 100[/tex]

In the problem, theoretical value is 5.80g and the measurement is 5.51g:

[tex]\frac{|5.80g-5.51g|}{5.80g} * 100[/tex]

% =

Answer:

Mg(s) + 2H^+(aq) ----> Mg^2+(aq) + H2(g)

Explanation:

We must first write the molecular equation, then obtain the complete ionic equation and subsequently get the net ionic equation as shown below;

Molecular reaction equation;

Mg(s) + 2HCl(aq) ------> MgCl2(aq) + H2(g)

Complete ionic equation;

Mg(s) + 2H^+(aq) + 2Cl^-(aq) ----> Mg^2+(aq) + 2Cl^-(aq) + H2(g)

Net ionic equation;

Mg(s) + 2H^+(aq) ----> Mg^2+(aq) + H2(g)

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:

Mass = 14.7 g

Explanation:

Given data:

Mass in g =?

Moles of K = 0.375 mol

Solution:

The Molar mass of potassium is 39.1 g/mol.

Formula

Number of moles = mass/molar mass

Mass = Number of moles × molar mass

Mass = 0.375 mol × 39.1 g/mol

Mass = 14.7 g

14.7 g K

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

0.375 mol K

Step 2: Identify Conversions

Molar Mass of K - 39.10 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

14.6625 g K ≈ 14.7 g K

Answer:

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

Answer:

Group 18 (VIIIa)

Explanation:

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.

Answer:

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

Answer:

1. Solid

2. Melting Point

3. liquid

4. Boiling point

5. gas

Answer:

Chemical to electrical to mechanical

Explanation:

The energy conversion going on in a battery - powered toy car is from chemical to electrical and then to mechanical energy.

According to the law of conservation of energy, energy is neither created nor destroyed but transformed from one form to another.

Answer:

Hallo

Explanation:

I took that question

Answer:

Most similar- Al

Least similar- F

Explanation:

From our knowledge of the periodic table, it is already clear that elements in the same group in the periodic table share certain similarities in physical and chemical properties.

Since Indium and aluminium belong to the same group in the periodic table, they will share a similarity in their boiling points.

Florine is a non metal, it is expected to be least similar to indium in physical properties.