What is the total pressure of mixture in which H2 and CO exerted 31mmHg and 510mmHg respectively​

Answer:

12 π electrons.

Explanation:

The structure of adenine can be seen below.

From the diagram, the total number of electrons in adenine is 70 electrons where 20 of them are core electrons.

Also from this same structure, we will notice that we have a total number of  12 π electrons.

i.e.

Each one of these five atoms described takes part in the π electron. These are N-1, N-3, N-7 which comprise of 2σ bonds, and also each of three N contributes one π electron.

Similarly, N-9 and N-10 contain 3σ bonds; Hence, each of them donates two electrons.

Thus;

5C = 5π electrons

3N = 3π electrons

2N = 4π electrons

     = 12π electrons

Answer:

They might explode. I am serious.

Answer:

0.102 M.

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M1) = 5.103 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (V2) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M1V1 = M2V2

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

The dilution of the NaOH results in the dilution of the molarity of the solution. The molarity of the final NaOH solution is 0.102 M.

Molarity is the concentration unit that defines the amount of solute present in a liter of solution.

The dilution of the solution results in a change in the molarity. It can be accessed as:

[tex]\rm M_1V_1=M_2V_2[/tex]

The initial molarity of the solution is, [tex]\rm M_1=5.103\;M[/tex]

The initial volume of the NaOH is, [tex]\rm V_1=5\;mL[/tex]

The final volume of the NaOH is, [tex]\rm V_2=250\;mL[/tex]

Substituting the values for the calculation of final molarity, [tex]\rm M_2[/tex]:

[tex]\rm 5.103\;\M\;\times\;5.0\;mL=M_2\;\times\;250\;mL\\\\M_2=\dfrac{5.103\;\M\;\times\;5.0\;mL}{250\;mL} \\\\M_2=0.102\;M[/tex]

The molarity of the NaOH solution after dilution is 0.102 M.

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Answer:

309 K

Explanation:

Step 1: Convert the pressure to atm

We will use the conversion factor 1 atm = 760 mmHg.

732 mmHg × 1 atm/760 mmHg = 0.963 atm

Step 2: Calculate the moles (n) of the ideal gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.963 atm × 8.4 L/0.0821 atm.L/mol.K × 298 K

n = 0.33 mol

Step 3: Calculate the temperature change

We will use the following expression.

Q = n × Cp × ΔT

ΔT = Q/n × Cp

ΔT = 75 J/0.33 mol × 20.79 J/mol.K

ΔT = 11 K

Step 4: Calculate the final temperature

T = 298 K + 11 K = 309 K

Since the forces equal to 0, it's balanced because the object isn't moving.

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Number of atoms in 1.4 mol of Phosphorus trifluoride (PF₃) : 8.428 x 10²³

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

1.4 mol of Phosphorus trifluoride (PF₃), number of atoms :

[tex]\tt N=1.4\times 6.02\times 10^{23}\\\\N=8.428\times 10^{23}[/tex]

Answer:

17.94 g

Explanation:

Answer:

In the given case, to get a blue flame, one should open the air inlet further. To increase the size of the flame one should open the gas regulator further. When one opens the air inlet, more amount of oxygen goes within, and thus, one can get a more intense form of blue flame.

When one opens the gas regulator more concentration of gas goes and the larger the size of flame one gets.

To regulate a bunsen burner follow this procedure

To get a blue flame, you should rotate the wheel  of the burner anticlockwise further to reduce the level of oxygen flow.

To increase the size of the flame, you should rotate the wheel clockwise further

The bunsen burner is an apparatus in the laboratory used to carry out experiment when heating of substance is required, hence it is used to add heat to substances, it has a wheel used to regulate the flow of air which is required for combustion, it has a hose which is used to tap gas to the burner

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Answer:

its DDDDDDDDDDDDDDDDDD

The structure that is least likely to exist is HS.

Sulfur is an element in group 16. Sulfur is a divalent element that has two lone pairs of electrons. This means that sulfur forms compounds in which it is bonded to two atoms or groups.

All the molecules listed can exist because they consists of structures in which sulfur is bonded to two atoms or groups. The only structure that can not exist is HS because it does not satisfy the valency of sulfur.

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Answer:

Atoms are not able to exist independently.

Explanation:

Brainliest pls

Answer:

Mega = 10^6

Milli = 10^-3

Explanation:

Mega is whatever times 10 to 6th power.

Milli is whatever times 10 to the -3rd power.

Answer:

answer d is correct that works

Orange is the color of Sodium

Answer:

NaClO; the contained elements ae Na, Cl, O.

a. molality-m=1.623

b. molarity-M=0.3

a. Boiling point

[tex]\tt \Delta Tb=Kb\times m[/tex]

Kb for ethanol=1.22 °C/m

Boiling point for ethanol = 78.4 °C

ΔTb=80.38-78.4=1.98

[tex]\tt 1.98=1.22\times m\\\\m=\dfrac{1.98}{1.22}=1.623[/tex]

b. osmotic pressure

[tex]\tt \pi =MRT[/tex]

π = 7.65 atm

R = 0.08205 L/atm K

T = 37 + 273 = 310 K

[tex]\tt M=\dfrac{\pi}{RT}=\dfrac{7.65}{0.08205\times 310}=0.3~M[/tex]

Answer:

a. PbSO4, Ksp = 1.7x10^-8.

Explanation:

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In this case, since the solubility product indicates how likely a solid is able to ionize and consequently dissolve in water, we can infer that the larger the solubility product Ksp, the more ions are able dissolve in water; therefore the proper answer goes with the largest Ksp, which is a. PbSO4, Ksp = 1.7x10^-8 since the power goes closer to 1 than the other options.

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Answer:

a. 167 mL b. 39.27 %

Explanation:

a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol  Al₂(SO₄)₃.

Now, we calculate the number of moles of Al in 45.0 g Al.

We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.

So n = 45.0 g/26.98 g/mol = 1.668 mol

Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol

So, we have 2.5 mol H₂SO₄.

Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.

V = n/C

= 2.5 mol/15.0 mol/L

= 0.167 L

= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.

b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃

Therefore 1.668 mol Al will produce x mol  Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol

So, we need to find the mass of 0.834 mol  Al₂(SO₄)₃. Now molar mass  Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.

Also number of moles of  Al₂(SO₄)₃, n = mass of  Al₂(SO₄)₃,m/molar mass  Al₂(SO₄)₃, M

n =m/M

So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g

% yield = Actual yield/theoretical yield × 100 %

Actual yield = 112 g, /theoretical yield = 285.2 g

So, % yield = 112 g/285.2 g × 100 %

= 0.3927 × 100 %

=  39.27 %

The volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum is 166mL and % yield of the reaction is 39.46%.

Moles of any substance will be calculated by using the below formula as:
n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is :

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

Moles of 45g of Al will be calculated as:

n = 45g / 27g/mol = 1.66 mole

From the stoichiometry of the reaction, it is clear that:

1.66 moles of Al = react with 3/2×1.66=2.49 moles of H₂SO₄

By using the formula of molarity we can calculate the volume of H₂SO₄ as:

M = n/V

V = (2.49) / (15) = 0.166L = 166mL

Again from the stoichiometry it is clear that:
1.66 moles of Al = produces 1/2×1.66= 0.83 moles of Al₂(SO₄)₃

Mass of 0.83 moles of Al₂(SO₄)₃ = (0.83mol)(341.96g/mol) = 283.82 g

Given actual yield of Al₂(SO₄)₃ = 112g

% yield will be calculated as:

Percent yield = (Actual yield/Theoretical yield) × 100

% yield = (112/283.82) × 100 = 39.46%

Hence required values are discussed above.

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Answer    1 mole = 6.02 × 1023 atoms

0.25 moles × (6.02 × 1023) = 1.5 × 1023 atoms Carbon

for the first one

Answer:

3grams

Explanation:

The reaction for the production of Magnesium dioxide will be

Mg + O2  → MgO

we have 5g of MgO (molar mass 40g)

no of moles of MgO = 5/40 = 0.125

Using unitary method we have

1 mole of Mg require 1 mole of MgO

0.125 Mole of MgO = 0.125mole of Mg

n = given mass /molar mass

0.125 = mass / molar mass

mass = 0.125* 24 = 3grams

Answer:

0.54M of Ba(OH)2

Explanation:

When Ba(OH)2 reacts with HCl, BaCl2 and H2O are produced as follows:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

The remanent hydroxide ion is because not all Ba(OH)2 reacts. Thus, we need to find moles of Ba(OH)2 that doesn't react and moles of Ba(OH)2 that reacts. The ratio between total moles and volume of the Ba(OH)2 solution = 0.050L is the molarity of the original solution

Moles of Ba(OH)2 that doesn't react:

50mL + 150mL = 0.200L * (0.12 mol OH- / L) = 0.024 moles OH-

2 moles of OH- are in 1 mole of Ba(OH)2:

0.024 moles OH- * (1mol Ba(OH)2 / 2 mol OH-) = 0.012 moles Ba(OH)2

Moles of Ba(OH)2 that react:

0.150L * (0.20mol / L) = 0.030 moles HCl

2 moles of HCl react per mole of Ba(OH)2:

0.030 moles HCl * (1mol Ba(OH)2 / 2 mol HCl) = 0.015 moles Ba(OH)2

Total moles:

0.012mol + 0.015mol = 0.027mol Ba(OH)2 in 50mL

0.027mol Ba(OH)2 / 0.0500L =

0.54M of Ba(OH)2

The following are not decomposition reactions :

B. CaO + H2O ⇒ Ca(OH)2

C. 2Mg + O2 ⇒ 2MgO

The decomposition reaction is a chemical reaction shows the decomposition  of a compound into its constituent elements or compounds

General formula :

AB⇒A+B

A. KCIO3 → KCI + O2

decomposition reaction

B. CaO + H2O ⇒ Ca(OH)2

synthesis/combination reaction

C. 2Mg + O2 ⇒ 2MgO

synthesis/combination reaction

D. 2NaOH ⇒ Na2O + H2O

decomposition reaction

Answer:

5.2 mol

Explanation:

Step 1: Given data

Step 2: Calculate the mole fraction of oxygen

We will use the following expression.

pO₂ = P × X(O₂)

X(O₂) = pO₂ / P

X(O₂) = 140 kPa / 530 kPa

X(O₂) = 0.264

Step 3: Calculate the moles of oxygen

We will use the definition of mole fraction of oxygen.

X(O₂) = n(O₂) / n

n(O₂) = X(O₂) × n

n(O₂) = 0.264 × 7.0 mol

n(O₂) = 1.8 mol

Step 4: Calculate the moles of nitrogen

The total number of moles is equal to the sum of moles of the individual gases.

n(O₂) + n(N₂) = n

n(N₂) = n - n(O₂)

n(N₂) = 7.0 mol - 1.8 mol

n(N₂) = 5.2 mol

Answer:  The volume occupied at an altitude of 20.0 km is 34.5289 L

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

where,

[tex]P_1[/tex] = initial pressure of gas = 101.325 kPa ( sea level)

[tex]P_2[/tex] = final pressure of gas = 10 kPa

[tex]V_1[/tex] = initial volume of gas = 3.40774 L

[tex]V_2[/tex] = final volume of gas = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]101.325\times 3.40774=10\times V_2[/tex]

[tex]V_2=34.5289L[/tex]

Therefore, the volume occupied at an altitude of 20.0 km is 34.5289 L

Answer:

Both sulphur and oxygen have two unshared pairs of electrons

Explanation:

The compound sulphur monoxide has the formula S=O. It is quite analogous to the diatomic molecules of the group 16 elements.

We must remember that each group element has six electrons in their outermost shell. Two of these are lone pairs leaving only two electrons available for bond formation.

These two electrons are used to form the sigma and pi bonds in S=O, leaving two lone pairs on each of sulphur and oxygen atoms.

Answer:

I play!!!!!!!!

Explanation:

Answer:

A and D

Explanation:

Answer:

no

Explanation:

Answer:

Water is heated as it travels through flat, glazed panels known as solar collectors, which are normally situated on the roof of the building. The heated water is then stored in an insulated storage tank.