Gino made a table to describe parts of the electromagnetic spectrum. A 3-column table with 4 rows. The first column labeled wave has entries ultraviolet, radio waves, infrared, x-rays. The second column labeled frequency has entries high, very low, low, high. The third column labeled Wavelength has entries long, very long, long, short. What mistake did Gino make? X-rays should have a low frequency and a long wavelength. Infrared light should have a high frequency, not a low frequency. Radio waves should have a very high frequency and a very short wavelength. Ultraviolet light should have a short wavelength, not a long wavelength.

Answer:

d and c

Explanation:

that is just the answer

Answer:

I cant see it that well but id be happy to help if u post a clearer picture!

Explanation:

Answer:

1. 11

2.45

3.54

5.11

Hope this helps if im not wrong

The IUPAC name for each of the compounds would be:

A. 2,6-Dimethyl octane

B. Octane

IUPAC naming is a system of naming organic compounds according to the rules set up by the International Union of Pure and Applied Chemistry.

According to these rules:

Applying these rules to the structures in the image, the IUPAC names would be as follows:

A. 2,6-Dimethyl octane

B. Octane

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Exothermic refers to chemical interactions that aerobic respiration. Combustion reactions release higher energy. Endothermic refers to atoms and molecules that either use or absorb reactive power.

A chemical process known as an endothermic releases energy as heat or light. It is an endothermic reaction's opposite. Chemical equation expressed as reactants + products + energy. An reaction mechanism is one in which electricity is given off as light or warmth.

A response is deemed to be exothermic if it produces heat while also undergoing a net decrease in basic enthalpy change. Samples include those type of combustion, iron rust, including water froze. Exothermic processes are those that discharge heat and energy into the surroundings.

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Answer:

1.) 0.619 M

2.) 6.34 L

3.) 12.7 moles

Explanation:

Part 1: To find the molarity, you first need to convert grams to moles using the molar mass.

Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol

Molar Mass (NaCl): 58.443 g/mol

143 grams NaCl               1 mole
-------------------------  x  ------------------------  =  2.45 moles NaCl
                                    58.443 grams

Molarity = moles / volume (L)

Molarity = 2.45 moles / 3.95 L

Molarity = 0.619

Part 2: To find the volume, you need to use the given moles and the previously calculated molarity.

Molarity = moles / volume (L)

0.619 M = 3.93 moles / volume

(0.619 M) x volume = 3.93 moles

volume = 6.34

Part 3: To find the moles, you need to use the given volume and the previously calculated molarity.

Molarity = moles / volume (L)

0.619 M = moles / 20.55 L

12.7 = moles

With few exceptions, cellular membranes — including plasma membranes and internal membranes — are made of glycerophospholipids, molecules composed of glycerol, a phosphate group, and two fatty acid chains. Glycerol is a three-carbon molecule that functions as the backbone of these membrane lipids.

Explanation:

the principal components of plasma membrane are lipids phospholipid and cholesterol proteins and carbohydrates

Answer:grass and wide open areas

Explanation: grassland- large open area of country covered with grass, especially one used for grazing.

Answer:

i needed the answer to this one too

Explanation:

The highest electric force exerted by charges -4 ×10⁻⁵ C and 3 ×10⁻⁵ C  placed 0.5 m apart is equal to 43.15 N.

The lowest electric force exerted by charges 3 ×10⁻⁵ C and 3 ×10⁻⁵ C  placed 1 m apart is equal to 8.10 N.

According to Coulomb’s law, the force of attraction between two charges is equal to the product of their charges and is inversely proportional to the square of the distance. This electric force applies along the line joining the two charges.

The magnitude of the electric force can be written as follows:

\(\displaystyle F = k\frac{q_1q_2}{r^2}\)

where k is constant proportionality = 8.99 × 10⁹ N.m²/C².

Given the charge on one point charge, q₁ =  4 ×10⁻⁵ C

The charge on the other point charge, q₂ = - 3 × 10⁻⁵C

The distance between these two charges, r = 0.5 m

The magnitude of electric force between the charges  will be:

\(\displaystyle F = 8.99\times 10^{9}\times \frac{4\times 10^{-5}\times 3\times 10^{-5}}{(0.5)^2}\)

F = 43.15 N

Given the charge on one point charge, q₁ =  3 ×10⁻⁵ C

The charge on the other point charge, q₂ = 3 × 10⁻⁵C

The distance between these two charges, r = 1 m

The magnitude of force between the charges will be:

\(\displaystyle F = 8.99\times 10^{9}\times \frac{3\times 10^{-5}\times 3\times 10^{-5}}{(1)^2}\)

F = 8.1 N

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Answer:

If there are more products than reactants, that means the reaction has shifted towards the left, which is the backward direction. If there are more reactants than products, that means the reaction has shifted towards the right, which is the forward direction.

Answer:

A chemical reaction is usually accompanied by easily observed physical effects, such as the emission of heat and light, the formation of a precipitate, the evolution of gas, or a color change.

The resulting pH of the solution is 1.74(approx).

To solve this problem, we need to use the concept of acid-base equilibrium and the pH scale. The addition of sulfuric acid will increase the concentration of H+ ions in the solution, which will shift the equilibrium of the \(H_2S\)/HS- system. We can use the following equation to calculate the pH of the resulting solution:

Ka = [H+][HS-]/\(H_2S\)]

where Ka is the acid dissociation constant for \(H_2S\), [\(H_2S\)], [HS-], and [H+] are the concentrations of the \(H_2S\), HS-, and H+ ions, respectively.

First, we need to calculate the initial concentrations of\(H_2S\) and HS- in the solution:

[\(H_2S\)] = 0.505 M

[HS-] = 0.505 M

Next, we need to calculate the amount of H+ ions added to the solution by 1 mL of concentrated sulfuric acid. To do this, we can use the following equation:

[H+] = (n/V) = (18.4 mol/L) x (1x\(10^{-3}\) L) = 1.84 x\(10^{-2}\)mol

where n is the amount of sulfuric acid added in moles, V is the volume of the solution in liters, and 18.4 mol/L is the concentration of the sulfuric acid.

Now, we can calculate the new concentrations of \(H_2S\), HS-, and H+ ions in the solution:

[\(H_2S\)] = [\(H_2S\)]0 - [H+] = 0.505 - 1.84x\(10^{-2}\)= 0.486 M

[HS-] = [HS-]0 + [H+] = 0.505 + 1.84x\(10^{-2}\) = 0.524 M

[H+] = 1.84 x \(10^{-2}\)M

Finally, we can use the equation for Ka to calculate the pH of the resulting solution:

Ka = 1.1 x \(10^{-7}\)

[H+] x [HS-]/[\(H_2S\)] = Ka

pH = -log[H+]

Substituting the values, we get:

(1.84 x \(10^{-2}\)) x (0.524)/(0.486) = 1.98 x \(10^{-2}\)

pH = -log(1.98 x \(10^{-2}\)) = 1.74(approx)

Therefore, the resulting pH of the solution is 1.74(approx)

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Answer:

4 CO + 8 H2 -> 4 CH3OH

Hope that it works

Answer:Approximate density of the high-leaded brass(alloy) =8.306g/cm³

Explanation:

The density of an  alloy is its mass (100g) divided by its volume

Therefore we have that the alloy (high-leaded brass) has a composition of

60.5 wt% Cu with density  8.94g/cm3

34.5 wt% Zn, wth densty  7.13g/cm3

5.0 wt% Pb with densty g/cm3

The total volume of the alloy will be the mass / density of ts composition given as :

60.5gCu/DCu + 34.5gZn/DZn + 5.0gPb/Dpb

= 60.5/8.94 + 34.5/7.13+ 5.0/11.35

= 6.76 cm³ + 4.838 cm³+0.4405 cm³ = 12.0385cm³

Approximate density of the high-leaded brass(alloy) = 100g/  12.0385cm³ =8.306g/cm³

Answer:

compound

Explanation:

compound is something that consists of two or more element.

The arrow in a chemical equation represents the direction of the reaction. It indicates the conversion of reactants into products. The arrow points from the reactant side to the product side, symbolizing the flow of the reaction.

The purpose of the arrow is to visually represent the chemical transformation occurring in the reaction. It shows the relationship between the reactants and products and the direction in which the reaction proceeds. The arrow implies that the reactant molecules are being rearranged and transformed into new substances with different properties.

Chemical equations are used to describe the stoichiometry and balance of reactions. The arrow helps convey this information by illustrating the overall process taking place. It serves as a crucial element in understanding the reaction's composition, reaction conditions, and the substances involved.

Furthermore, the arrow also implies that the reaction can occur in both directions. In reversible reactions, the arrow can be represented as a double-headed arrow, indicating that the reaction can proceed in either direction depending on the conditions.

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A)  the addition of \(H_3O\)+ or OH- ions would have opposing effects on the buffer solution.

b) \(H_3O\)+ would favor the formation of acetic acid, while adding OH- would favor the formation of acetate ions.

c) The exact magnitude of the changes in concentrations depends on the initial concentrations of \(CH_3COOH\) and \(CH_3COONa\), as well as the specific amount of \(H_3O\)+ or OH- added.

To determine the effect of adding \(H_3O\)+ and OH- on the given buffer solution, we need to consider the ionization of acetic acid (\(CH_3COOH\)) and the dissociation of its sodium salt (CH3COONa). Let's analyze each scenario separately:

A) Buffer solution consisting of 0.5 M\(CH_3COOH\) and 0.5 M \(CH_3COONa\):

When acetic acid (CH3COOH) and its sodium salt (\(CH_3COONa\)) are present together in a solution, they form a buffer system. Acetic acid partially ionizes in water, releasing \(H_3O\)+ ions, while sodium acetate dissociates into Na+ and \(CH_3COO\)- ions.

Adding \(H_3O\)+:

The \(H_3O\)+ ions would react with the acetate ions (CH3COO-) to form undissociated acetic acid (\(CH_3COOH\)) through the following reaction:

\(H_3O\)+ + \(CH_3COO\)- ⇌ \(CH_3COOH\) + H2O

The addition of H3O+ would shift the equilibrium to the left, promoting the formation of more acetic acid and decreasing the concentration of acetate ions.

Adding OH-:

The OH- ions would react with the acetic acid (\(CH_3COOH\) to form water and acetate ions (CH3COO-) through the following reaction:

OH- + \(CH_3COOH\) ⇌ \(CH_3COO\)- + H2O

The addition of OH- would shift the equilibrium to the right, consuming acetic acid and increasing the concentration of acetate ions.

B) After adding 0.02 mol of NaOH to 1.0 L of the buffer solution:

When solid NaOH is added to the buffer solution, it dissociates completely in water to form Na+ and OH- ions.

NaOH dissociation:

NaOH → Na+ + OH-

The OH- ions formed would react with acetic acid according to the reaction mentioned in Scenario A (2), increasing the concentration of acetate ions and consuming acetic acid.

C) After adding 0.02 mol of HCl to 1.0 L of the buffer solution:

When HCl is added to the buffer solution, it dissociates completely in water to form \(H_3O\)+ and Cl- ions.

HCl dissociation:

HCl → \(H_3O\)+ + Cl-

The \(H_3O\)+ ions formed would react with acetate ions (\(CH_3COO\)-) according to the reaction mentioned in Scenario A (1), forming more undissociated acetic acid and decreasing the concentration of acetate ions.

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Mass of B₂ = 18.75 g

Given

Element A and element B

Required

The reaction

mass of B₂

Solution

The balanced equation :

2A(s) + B₂(g) ⇒2AB(s)

mol AB :

= Molarity x Volume

= 2.5 M x 0.25 L

= 0.625

From the equation, mol ratio of B₂ : AB = 1 : 2, so mol B₂ :

= 1/2 x mol AB

= 1/2 x 0.625

= 0.3125

Mass B₂ (Molar mass = 2x30 g/mol=60 g/mol)

= mol x molar mass

= 0.3125 x 60 g/mol

= 18.75 g

Answer:

the answer is 373

Explanation:

K=°C + 273

K=100+273=373

250 L of 5.0% w/v glucose solution can be prepared using 125 g of glucose.

w/v = [ mass of solute (g) / volume of solution (mL) ] × 100

5.0 % w/v = [ 125 g / volume of solution (mL) ] × 100

volume of solution (mL) = [ 125 g / 5.0 % w/v ] x 100

volume of solution (mL) = [ 125 g / (5.0 / 100) ] x 100

volume of solution (mL) = [ 125 g / 0.05 ] x 100

Answer:

\(296.05\ \text{K}\)

Explanation:

\(m_w\) = Mass of water = 100 g

\(c_w\) = Specific heat of water = \(4.184\ \text{J/g K}\)

\(m_c\) = Mass of copper = 20 g

\(c_c\) = Specific heat of copper = \(0.385\ \text{J/g K}\)

\(\Delta T_w\) = Temperature change in water = \((T-295)\)

\(\Delta T_c\) = Temperature change in cooper = \((353-T)\)

T = Final temperature of the system

The heat balance of the system is given by

\(m_wc_w\Delta T_w=m_cc_c\Delta T_c\\\Rightarrow 100\times 4.184\times (T-295)=20\times 0.385\times (353-T)\\\Rightarrow 418400\left(T-295\right)=7700\left(353-T\right)\\\Rightarrow 418400T-123428000=2718100-7700T\\\Rightarrow T=\frac{1261461}{4261}\\\Rightarrow T=296.05\ \text{K}\)

The final temperature of the water is \(296.05\ \text{K}\).

The final temperature of the water when placed in a calorimeter is 296.05K

HOW TO CALCULATE FINAL TEMPERATURE:

Where;

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Answer:

Ca(C2H3O2)2 has 30.41% carbon by volume

Explanation:

Answer:

d

Explanation:

higher no of particles are in CaCl2 solution

The depression in freezing point, elevation in boiling point, etc. are known as colligative particles. Here the CaCl2 solution has the higher boiling point and the NaCl solution has the lower freezing point. The correct option is D.

The properties of a dilute solution which depend only on the number of solute particles present in a given amount of solvent, and not in any way upon the nature of the solute particles are called colligative properties.

The elevation in boiling point is:

ΔTb = i × Kb × m

Here 'i' is the Van't Hoff factor, Kb is constant and 'm' is the molality.

The boiling point depends on the number of ions present in the electrolyte and the concentration. The number of ions in CaCl₂ is 3 whereas the number of ions in NaCl is 2.

ΔTb of CaCl₂ = 3 × 0.25 = 0.75

ΔTb of NaCl = 2 × 0.25 = 0.5

Similarly the freezing point depends on the number of particles of the electrolyte.

So CaCl₂ has boiling point and NaCl has low freezing point.

Thus the correct option is D.

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Answer:

Sb

Explanation:

The periodic trend for atomic radius is that it decreases from left to right and increases from top to bottom, therefore the elements with the larger atomic radius will be the ones which are closest to the bottom left corner of the periodic  table. Since all of these elements are in the same group, the one with the largest atomic radius will be the one at the "bottom", and that is Sb.

The mass concentration of sand in the ethanol solution is 0.299 g/dm³.

To find the concentration in grams per cubic decimeter (g/dm³), we first need to convert the volume from liters to cubic decimeters (dm³). Since 1 liter is equal to 1 cubic decimeter, we can directly convert the volume.

Given:

Mass of sand = 0.2 g

Volume of ethanol = two-thirds liter

Converting volume to dm³:

1 liter = 1 cubic decimeter

two-thirds liter = (2/3) cubic decimeter = 0.67 dm³ (rounded to two decimal places)

Now we can calculate the concentration in g/dm³ by dividing the mass of sand by the volume in dm³:

Concentration = Mass / Volume

Concentration = 0.2 g / 0.67 dm³

Concentration ≈ 0.299 g/dm³ (rounded to three decimal places)

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Answer:

its D:)

Explanation:

NH3 will dissociate into N2 and H2.

Given the equation of the reaction; N2 (g) + 3H2 (g) -----> 2NH3(g)

We can calculate the reaction quotient as follows;

Q = [NH3]^2/[N2] [H2]^3

From the question;

[NH3] = 0.0100 M

[N2] = 0.0200 M

[H2] = 0.0600 M

Substituting values;

Q = [0.0100 M]^2/[0.0200 M] [0.0600 M]^3

Q = 23.1

Since Q > Kc, NH3 will dissociate into N2 and H2.

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50.9 grammes of Nickel (III) oxide are converted into 36.2 grammes of Nickel.

By extension of this definition, the mass of any material represented in atomic mass units is numerically equivalent to the molar mass of that substance in grammes per mole. For instance, an oxygen atom has an atomic mass of 16.00 amu, which translates to a molar mass of 16.00 g/mol.

2Nickel (III) oxide(s) ⟶ 4Nickel(s) + 3Ocygen(g)

The balanced equation shows that 2 moles of Nickel (III) oxide reacts to form 4 moles of Nickel.

Next, we need to calculate the number of moles of Nickel (III) oxide:

molar mass of Nickel (III) oxide = 165.38 g/mol

moles of Nickel (III) oxide = mass/molar mass = 50.9 g/165.38 g/mol = 0.308 moles

Using the mole ratio from the balanced equation, we can calculate the moles of Nickel formed:

moles of Nickel = 0.308 moles Nickel (III) oxide × (4 moles Nickel /2 moles Nickel (III) oxide) = 0.616 moles Nickel

Finally, we can calculate the mass of Nickel formed:

molar mass of Nickel = 58.69 g/mol

mass of Nickel = moles of Nickel × molar mass of Nickel = 0.616 moles × 58.69 g/mol = 36.2 g

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Answer:

nonmtals:

Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process. It is also used for manufacturing of steel and provides high temperature in metal fabrication process. ...

Nonmetals used in fertilizers: Fertilizers contain nitrogen. It helps in plant growth. ...

Nonmetals used in crackers

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Seven non-metals exist under standard conditions as diatomic molecules: H2(g)

Explanation: