A 12 inch object is placed 30 inches in front of a plane mirror

Answer:

The answer is the first option 0.37 moles

Explanation:

the ideal gas law is PV = nRT, so in the terms of n,

n = PV/RT (divide both sides by RT to let n stand alone)

given: P = 700 kpa, V = 2L, T = 450 K , R = 8.31

n = (700 * 2) / (8.31 * 450)

= 1400/3739.5

= 0.34

Answer:

125 ohms

Explanation:

since they're parallel we do

1/total resistance = (1/250 ohms) + (1/250 ohms)

then you get 2/250 ohms

then you flip it so it's equal to total resistance/1

250÷2= 125

Answer:

0.190L of hydrogen may be produced by the reaction.

Explanation:

Our reaction is:

3Mg + 2H₃PO₄ →  Mg₃(PO₄)₂ + 3H₂

We need to determine the limting reactant. Let's find out the moles of each:

5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg

55.23 g . 1 mol / 97.97 g = 0.563 moles of acid

2 moles of acid react to 3 moles of Mg

0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg

Definetely the limting reactant is Mg.

As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen

Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂

At STP, 1 mol of any gas occupies 22.4L

0.00857 mol . 22.4L / 1mol = 0.190L

Answer:

a. 408g de NH3 son necesarios

b. 2.42x10²³ moléculas de agua se obtienen

Explanation:

a. Basados en la reacción, 2 moles de NH3 producen una mol de (NH2)2CO. Para resolver esta pregunta debemos convertir la masa de (NH2)2CO a moles. Estas moles a moles de NH3 y su respectiva masa:

Moles (NH2)2CO -Masa molar: 60.06g/mol-

720g * (1mol / 60.06g) = 11.99 moles (NH2)2CO

Moles NH3:

11.99 moles (NH2)2CO * (2mol NH3 / 1mol (NH2)2CO) = 23.98 moles NH3

Masa NH3 -17.031g/mol-:

23.98 moles NH3 * (17.031g / mol) = 408g de NH3 son necesarios

b. 1mol de CO2 produce  1mol de agua. Se debe convertir el volumen a moles usando PV = nRT. Estas moles = Moles de agua. 1mol = 6.022x10²³ moléculas:

Asumiendo STP:

PV = nRT; PV / RT = n

Donde P = 1atm a STP

V = 9L

R = 0.082atmL/molK

T = 273.15K a STP

1atm*9L / 0.082atmL/molK*273.15K = n

n = 0.402 moles CO2 = Moles H2O

Moléculas:

0.402 moles agua * (6.022x10²³ moléculas / 1mol) =

Answer:

B. Cation

Explanation:

As an electron is removed from the atom it is then converted into a positively charged ion known as a Cation.

Hope this helps u

Answer:

The answer is c

Answer:

3132 is the answer

Explanation:

3BaCl2 + Al2S3 → 3BaS + 2AlCl3

Answer:

Explanation:

Hope this helps u !!

Answer:

weak ion attractions in the table salt

Answer:

121 aprox.

Explanation:

mass/molar mass = Q/CF

where mass=?

molar mass of Al=27

Q=It = 15×24×60×60

C=3

F=96500

Answer:

the correct answer is zinc

Answer:

[tex]\boxed {\boxed {\sf 81.4 \ g\ KOH}}[/tex]

Explanation:

We are asked to convert 1.45 moles of potassium hydroxide (KOH)  to grams.

The molar mass is the mass in grams of 1 mole of a substance. The values are found on the Periodic Table. They are the same as the atomic masses but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We have the compound KOH. Look up the molar masses of the individual elements.

The formula for the compound has no subscripts, so there is 1 mole of each element in 1 mole of the compound. We can simply add the molar masses.

Based on the molar mass, there are 56.105 grams of KOH in 1 mole. Create a ratio using this information.

[tex]\frac {56.105 \ g\ KOH}{ 1 \ mol \ KOH}[/tex]

We are converting 1.45 moles of KOH to grams, so we multiply by this value.

[tex]1.45 \ mol \ KOH *\frac {56.105 \ g\ KOH}{ 1 \ mol \ KOH}[/tex]

The units of moles of KOH cancel.

[tex]1.45\ *\frac {56.105 \ g\ KOH}{ 1 }[/tex]

[tex]81.35225 \ g\ KOH[/tex]

The original measurement of moles has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the tenths place. The 5 in the hundredths place (81.35225) tells us to round the 3 up to a 4.

[tex]81.4 \ g\ KOH[/tex]

Approximately 81.4 grams of potassium hydroxide are represented by 1.45 moles.

Answer:

2.34 %

Explanation:

Since the density of the Jello, ρ = 1.14 g/mL and ρ = m/v where m = mass of jello and v = volume of jello = 475 mL.

So, m = ρv

substituting the values of the variables into the equation, we have

m = ρv

m = 1.14 g/mL × 475 mL = 541.5 g

Since we have 13 g of sugar in the jello, the total mass present is 13 g + 541.5 g = 554.5 g

So, the percentage by mass of sugar present % m/m = mass of sugar present/total mass × 100 %

= 13 g/554.5 g × 100 %

= 0.0234 × 100 %

= 2.34 %

So, the percentage by mass of sugar present % m/m = 2.34 %

Answer:

hola

Explanation:because si

Answer:

CO2 it is an element.

Hope it helps:)

Answer:

Ozone layer depletion is the thinning of the ozone layer present in the upper atmosphere. This happens when the chlorine and bromine atoms in the atmosphere come in contact with ozone and destroy the ozone molecules. One chlorine can destroy 100,000 molecules of ozone. It is destroyed more quickly than it is created

Answer:

Ozone depletion, gradual thinning of Earth’s ozone layer in the upper atmosphere caused by the release of chemical compounds containing gaseous chlorine or bromine from industry and other human activities. The thinning is most pronounced in the polar regions, especially over Antarctica. Ozone depletion is a major environmental problem because it increases the amount of ultraviolet (UV) radiation that reaches Earth’s surface, which increases the rate of skin cancer, eye cataracts, and genetic and immune system damage. The Montreal Protocol, ratified in 1987, was the first of several comprehensive international agreements enacted to halt the production and use of ozone-depleting chemicals. As a result of continued international cooperation on this issue, the ozone layer is expected to recover over time.

Answer:D

Explanation:

Answer:

See explanation

Explanation:

The interior of the atom is composed of electrons which move around in orbits and the nucleus. The nucleus contains protons and neutrons. The electrons are negatively charged while the protons are positively charged, the neutrons have no charge at all.

The atom is held together by electrostatic interaction between the electrons and protons in the nucleus. The protons and neutrons account for most of the mass of the atom.

The greatest wonder that I will like to really understand in the interior of the atom is the nature of the interaction between the neutrons and protons in the atomic nucleus which holds the nucleons together. The neutrons have no charge while the protons are positively charged. What sort of interaction could exist between an uncharged particle and a charged particle? That is what I will like to understand better as a result of my travels into the interior of the atom.

Answer:

help

Explanation:

Answer:

gas ; methane , ethane  

liquid ; octane , hexane

Explanation;

i cant find solid dear

pls help anyone with an example of a  solid alkanes

Answer:

it should be powdered zinc with high concentration of sulfuric acid

Answer:

The energy absorbed in the first move is greater than the energy released in the second move

Explanation:

Electrons require (absorb) energy to move to a higher energy level when there is a large external heat source, the presence of an electric field or by colliding with other electrons

And the amount of energy absorbed by the electron is exactly equal to the change in the energy state between the initial energy level of the electron and the destination energy level

Therefore, given that the energy level of the electron at level 2 is higher than the energy level of the electron when at level 1, we have;

The difference in the energy level between level 4 and level 1 is greater than the difference in the energy level between level 4 and level 2 and more energy is absorbed and therefore, released when the electron moves from level 1 to level 4 than when the electron drops from level 4 to level 2.

The most likely result is that 'the energy absorbed in the first move is greater than the energy released in the second move'.

Answer:

D a plank floating on water

Explanation:

Its not a mixture

Answer:

STP is 101.325 kPa pressure and 273K temperature

Also, at STP, 1 mol of an ideal gas = 22.4 L

Since the question states that 22.4 L are needed at STP, we know that ONE MOLE is needed.  Now, the question is, how many moles are actually present?

V = 25.0 L

P = 101.5 kPa

T = 373K

n = ?

R = 8.31 L-kPa/Kmol

Solving the ideal gas law for n (moles) and comparing that to 1 mole, we have...

n = PV/RT = (101.5)(25.0)/(8.31)(373)

n = 0.82 moles of gas.  THIS IS NOT ENOUGH as it is less than the ONE mole required.

Explanation:

Explanation:

The haem is in turn split into iron , which forms chemical compounds as part of the blood iron pool available for future hemoglobin synthesis...

hope this help !!

Answer:

The given molecule is KCN.

Is it a ternary ionic compound?

Explanation:

A ternary ionic compound is one that consists of at least one polyatomic ion/radical.

In the given molecule,

the ions present are:

K+ and CN-.

CN- is a polyatomic acidic radical.

Hence, the given molecule is ternary ionic compound.

Answer:

Explanation:

The answer is C Enthaply and entropy because temperature plays a big role in reactions

Answer:

c

Explanation:

a p e x :)

Answer:

false

Explanation:

it happens in light-independent reactions

Hello. You have not presented the image that presents the two elements mentioned in the question. This makes it impossible for your question to be answered. However, I will try to help you as best I can.

To say whether the compounds obey the law of multiple proportions, you should observe whether one of the elements, formed from the compounds, maintains a fixed mass, while the other element presents the mass in a varied amount of small, whole numbers, spread across the formed compounds .

This is because the law of multiple proportions states that an element must have a fixed mass when reacting with another element to create compounds. This reaction will allow this element to keep the mass fixed, while the other element will generate different compounds, where each one presents a part of the mass of the forming element, in small and whole numbers.