I have an unknown volume of gas held at a temperature of 115 K in a container with a pressure of 60atm. If by increasing the temperature to 225 K and decreasing the pressure to 30. atm causes the volume of the gas to be 29 liters, how many liters of gas did I start with?
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Answer: The question is not clear

Explanation:

The chemical formula of water is H2O, and the chemical formula of carbon dioxide is CO2. This means that water has 2 atoms of hydrogen and 1 atom of oxygen, which carbon dioxide has 1 atom of carbon and 2 atoms of oxygen. The only similar atom that the two molecules have in common is oxygen. CO2 has one more atom of oxygen than H2O. Hope this helps!

H2O is water. CO2 is CD.

Answer:

a. True.

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]

[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

A. True

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

D. beta particle

Explanation:

Pu has 94 protons which is only one more than Np (93), that means beta decay. Atomic mass stays the same in beta decay. Beta is very small particle so atomic mass stays at 239.

The given reaction is an example of beta decay where the atomic number of the nuclei increases by one unit and mass number remains unchanged. Therefore, the particle X is a beta particle or an electron.

Heavy unstable isotopes of atoms undergoes nuclear decay by the emission of charged particles such as alpha or beta particle. In alpha decay, the helium nuclei is emitted and in beta decay, electrons are emitted.

In alpha decay, the mass number of the nuclei decreases by 4 units and atomic number by 2 units. In beta decay, the mass number does not change but the atomic number increases by one unit.

Neptunium undergoes beta decay and forms the plutonium nuclei. Thus X is a beta particle. The reaction is written as follows:

[tex]\rm _{93} ^{239}Np \rightarrow _{94}^{239}Pu + _{-1}^{0}e[/tex]

To find more on beta decay, refer here:

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Answer:

35

Explanation:

is the answer for your question

Nitrogen gained 4 electrons.

Because Nitrogen's redox number went from +6 to +2, it must have gained 4 electrons (-4) in order to achieve this number. Thus, Nitrogen is reduced.

The correct answer is C. square planar

According to the Valence Shell Electron Pair Repulsion Theory(VSEPR), The shape of a molecule depends on the number of electron pairs in the molecule.

VSEPR theory was first coined by Gillespie  and Nyhlom in 1957 as an improvement over the Sidgwick - Powell theory.

According to this theory, the shape of a molecule is determined by the number of electron pairs that surround the valence shell of the central atom in the molecule.  The electron pairs are positioned as far apart in space as possible to minimize repulsion of electron pairs.

However, the presence of lone pairs distorts the shape anticipated for the molecule on the basis of VSEPR.

For a molecule having six electron pairs, an octahedral geometry is expected(electron domain geometry). However, the presence of two lone pairs which are positioned at opposite side of the four single bonds leads to an observed square planar molecular geometry.

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Answer:

square planar

Explanation:

Answer:

79 g/mol

Explanation:

Mass of unknown metal deposited = 3.137 g - 1.4 g = 1.737 g

Number of moles of metal deposited = 0.022 moles

Since;

Number of moles = reacting mass/molar mass

Molar mass = reacting mass/number of moles

Molar mass = 1.737 g/0.022 moles

Molar mass= 79 g/mol

Answer: 32 electrons

Explanation:

Answer:

dsgsdfd

Explanation:

Answer:

D. An alkene ​

Explanation:

because Ethane is C2H4

Answer:

It's a alkANE. C.

Explanation:

The easiest way to memorize this is to look at the endings. Substances that end in -ANE are alkANEs. Substances that end in -ENE are alkENEs. Substances that end in -YNE are alkYNEs.

Answer:

[tex]n_1=4[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=489.2 nm =>4.86*10^{-7}[/tex]

nf level= Balmer series

nf level= 2

Generally the equation for Wavelength is mathematically given by

[tex]\frac{1}{\lambda}=R[\frac{1}{nf^2}-\frac{1}{n_1^2}][/tex]

Where

[tex]R=Rydberg Constant[/tex]

[tex]R=1.097*10^7[/tex]

Therefore

[tex]\frac{1}{4.86*10^{-7}}=1.097*10^7[\frac{1}{2^2}-\frac{1}{n_1^2}][/tex]

[tex]n_1=4.0021[/tex]

[tex]n_1=4[/tex]

Explanation:

here's the answer to the question

Answer:

[tex]C_t=0.165M[/tex]

Explanation:

From the question we are told that:

Slope [tex]K=0.056 M-1 s -1[/tex]

initial Concentration [tex]C_1=2.2M[/tex]

Time [tex]t=100[/tex]

Generally the equation for Raw law is mathematically given by

[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]

[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]

[tex]C_t=0.165M[/tex]

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]

Substitute values in the above equation:

[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]

Therefore, after 100 seconds the concentration is 0.165 M.

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Answer:

5g/cm3

Explanation:

firstly convert the litres and kilograms to grams and centimeters.

1l is equivalent to 1000cm3

0.58×1000

580cm3

and 1kg is equivalent to 1000g

2.9×1000

2900

then find the density by using the formula

density=mass/volume

=2900g/580cm3

=5g/cm3

I hope this helps

Answer:

B.bonds are broken and new bonds are formed

Answer:A) V = 16.892 ml

Explanation:

M1 * V1 = M2 * V2

14.8 M * V1 =0.250 M * 1000 ml

V1 = 16.892 ml

a. The volume of 16.89 milliliters of the stock solution of 14.8 M  should be diluted to make 1000.0 mL of 0.250 M.

b. The concentration of the final solution is 0.296 M.

The concentration or the volume of the concentrated or dilute solution can be calculated by using the equation:

M₁V₁ = M₂V₂

where M₁ and V₁ are the concentration and volume of the concentrated solution respectively and M₂ and V₂ are the concentration and volume of the dilute solution.

A stock solution is a solution that has a high concentration and that will be diluted to a low concentration by the addition of water in it.

Given, a stock solution of concentration, M₁ = 14.8 M

The concentration of the diluted solution, M₂ = 0.250 M

The volume of diluted solution, V₂  = 1000ml

Substitute the value of the molarity and volume in equation (1):

(14.8)× (V₁) = (1000) × (0.250)

V₁ = 16.89 ml

Similarly, for part (b): M₁ = 14.8 M, V₁ = 10 ml and V₂  = 0.5L = 500 ml

(14.8)× (10) = (500) × (M₂)

M₂ = 0.296 M

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Answer:

0.074 moles

Explanation:

For every mole (of any element), there are 6.022 x 10^23 atoms.

There are 4.45 x 10^22 atoms of iron.

To find the moles we divide the number of atoms by Avogadro's number

4.45 x 10^22 / 6.022 x 10^23 = 0.0738957

Don't forget sig figs

Answer:

[tex]\boxed {\boxed {\sf 0.0739 \ mol \ Fe}}[/tex]

Explanation:

We are asked to convert a number of iron atoms to moles of iron.  

We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of iron. There are 6.022 ×10²³ atoms of iron in 1 mole of iron.  

We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.

[tex]\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]

Since we are converting 4.45 × 10²² atoms of iron to moles, we multiply the ratio by that value.

[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]

Flip the ratio. The value is the same, but it allows us to cancel the units of atoms of iron.

[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {1 \ mol \ Fe}{6.022 \times 10^{23} \ atoms \ Fe}[/tex]

[tex]4.45 \ \times 10^{22} *\frac {1 \ mol \ Fe}{6.022 \times 10^{23}}[/tex]

Condense into 1 fraction.

[tex]\frac {4.45 \ \times 10^{22} }{6.022 \times 10^{23}} \ mol \ Fe[/tex]

[tex]0.07389571571 \ mol \ Fe[/tex]

The original measurement of atoms ( 4.45 × 10²²) has 3 significaint figures, so our answer must have the same. For the number we calculated, that is the ten-thousandths place. The 9 to the right of this place (0.07389571571) tells us to round the 8 up to a 9.

[tex]0.0739 \ mol \ Fe[/tex]

There are approximately 0.0739 moles of iron in 4.45 × 10²² atoms of iron.  

the effects are: temperature rises, water shortages, and increased fire threats

Explanation:

try 96 if that's not right let me know and I'll try to fix it

Answer:24

Explanation:

Answer:

"They established the laws of planetary motion"

Explanation:

Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."

Answer:

Answer:

As a giant star.

Explanation:

A

Answer:

I think carbon dioxide is right answer

Answer:

It is prepared small amounts of hydrogen cloride for uses in the lab.

It can be  "generated in an HCl generator by dehydrating hydrochloric acid with either sulfuric acid or anhydrous calcium chloride."

Answer:

Explanation:

Start with a balanced equation.

2H2 + O2 → 2H2O

Calculate mole H2 using the formula: n = m/M, where:

n = mole

m = mass (g)

M = molar mass (g/mol)

Calculate molar mass of H2.

M H2 = 2 × 1.008 g/mol = 2.016 g/mol

Calculate moles H2.

n H2 = 4.16 g H2/2.016 g/mol = 2.063 mol H2

Calculate moles H2O by multiplying moles H2 by the mole ratio between H2O and H2 from the balanced equation, so that moles H2 cancel.

2.063 mol H2 × (2 mol H2O/2 mol H2) = 2.063 mol H2O

The mass of water will be calculated by rearranging the n = m/M formula to isolate m;

m = n × M

Calculate the molar mass H2O.

M H2O = (2 × 1.008 g/mol) + (1 × 15.999 g/mol) = 18.015 g/mol

Calculate the mass H2O.

m = n × M = 2.063 mol H2O × 18.015 g/mol = 37.2 g H2O

4.16 g H2 with excess O2 will produce 37.2 g H2O.

Answer:

the number is 7

Explanation:

"Three times" means multiply by 3

"Difference" means subtract

"Sum" means add

3(x - 7) = 23 - (3x + 2)

3x - 21 = 23 - 3x - 2

3x - 21 = 21 - 3x

6x = 42

x = 7

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

[tex]\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}[/tex]

Explanation:

We are asked to find how many molecules of methane are in 14.8 grams of the substance.

First, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.

We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).

Check the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.

Now we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.

[tex]\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Since we are converting 14.8 grams of methane to moles, we multiply by this value.

[tex]14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Flip the ratio so the units of grams of methane cancel.

[tex]14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }[/tex]

[tex]14.8 *\frac{ 1 \ mol \ CH_4} {16.043}[/tex]

[tex]\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4[/tex]

Next, we convert moles to molecules. We use Avogadro's Number or 6.022  × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.

[tex]\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

Multiply by the number of moles we calculated.

[tex]0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

The units of moles of methane cancel.

[tex]0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }[/tex]

[tex]5.55541981 \times 10^{23} \ molecules \ CH_4[/tex]

The original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.

[tex]5.56 \times 10^{23} \ molecules \ CH_4[/tex]

14.8 grams of methane is equal to approximately 5.56 × 10²³ molecules of methane.