For the following reaction, 61.6 grams of bromine are allowed to react with 25.5 grams of chlorine gas. bromine (g) + chlorine (g) bromine monochloride (g) What is the maximum amount of bromine monochloride that can be formed? grams

Answer:

On moving from left to right in a period the electropositive character decreases as the tendency to lose electrons decreases.

Explanation:

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:

Consequence

Answer:

74.3g of methanol were introduced into the vessel

Explanation:

In the equilibrium:

CH₃OH(g) ⇄ CO(g) + 2H₂(g)

Kc is defined as the ratio between concentrations in equilibrium of :

Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]

Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:

[CH₃OH] = ? - X

[CO] = X

[H₂] = 2X

Where ? is the initial concentration of methanol

As [H₂] = 2X = 0.426M; X = 0.213M

[CH₃OH] = ? - 0.213M

[CO] = 0.213M

[H₂] = 0.426M

Replacing in Kc to solve equilibrium concentration of methanol:

6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]

[CH₃OH] = 0.560

As:

[CH₃OH] = ? - 0.213M = 0.560M

? = 0.773M

0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:

3.00L * (0.773 mol / L) = 2.319 moles methanol.

Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:

2.319 moles * (32.04g / mol) =

Answer:

Silver.

Explanation:

Hello,

In this case, for the redox reaction:

[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]

We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.

Best regards.

The oxidizing agent in the redox reaction represented by the following cell notation is Silver.

The redox reaction is

Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)

here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand,  the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.

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Answer:

b) dime

Explanation:

a dime is approximately 2.2g

half of this is 1.1g, which can be rounded down to one gram.

hope this helps

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

The entropy change in the surroundings associated with this reaction occurring at 25 degree C is calculated as ΔS = -ΔH/T J/K.

Entropy is a quantity which gives idea about the randomness or arrangement of atoms or molecules present in any sample.

Entropy change will be calculated as:
ΔS = -ΔH/T, where

ΔH = chnage in enthalpy (J/mole)

T = temperature (K)

So to calculate the entropy change first we have to know about the value of enthalpy in joules and then divide it by the temperature.

Hence the unit of entropy is joule per kelvin.

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Answer:

[tex]x_{et}=0.6068[/tex]

Explanation:

Hello,

In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:

[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]

Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)

[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]

Therefore, the mole fraction turns out:

[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]

Best regards.

Answer:

Mass = 11.688g

Explanation:

Volume = 2.00L

Molar concentration = 0.100M

Mass = ?

These quantities are relatted by the following equation;

Conc = Number of moles / volume

Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol

Number of moles = Mass / Molar mass

Mass = Number of moles * Molar mass

Mass = 0.2mol * 58.44g/mol

Mass = 11.688g

Answer:

The correct answer is 1.60 Liters.

Explanation:

The given reaction:

CO (g) + H₂(g) ⇔ CH₃OH (g)

Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.

It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.

As 22.4 L at standard temperature and pressure is equivalent to 1 mole.

Therefore, 1 L at STP will be, 1/22.4 mole

Now 3.20 L at STP will be,

= 1/22.4 × 3.20

= 0.1428 mole

And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH

Now, 0.1428 mole of H₂ will give,

= 0.1428/2 = 0.071 mole of CH₃OH

= 0.071 × 22.4 = 1.60 L

The volume, in liters, of CH₃OH gas formed is 1.60 L

From the question,

We are to determine the volume of CH₃OH formed

The given chemical equation for the reaction is

CO(g)+ H₂(g) → CH₃OH

The balanced chemical equation for the reaction is

CO(g)+ 2H₂(g) → CH₃OH

This means

1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

x mole of the H₂ gas will have a volume of 3.20 L at STP

x = [tex]\frac{3.20 \times 1}{22.4}[/tex]

x = 0.142857 mole

∴ The number of mole of H₂ present is 0.142857 mole

Since

2 moles of H₂ reacts to produce 1 mole of CH₃OH

Then,

0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH

[tex]\frac{0.142857}{2} = 0.0714285[/tex]

∴ The number of moles of CH₃OH produced = 0.0714285 mole

Now, for the volume of CH₃OH formed

Since

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285  at STP

22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L

Hence, the volume of CH₃OH gas formed is 1.60 L

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Answer:

Below

Explanation:

Let n be the quantity of matter in the Calcium Bromide

● n = m/ M

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)

M = 40.1 + 2×79.9

● 0.422 = m/ (40.1+2×79.9)

●0.422 = m/ 199.9

● m = 0.422 × 199.9

● m = 84.35 g wich is 88.4 g approximatively

88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.

Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .

To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,

Let n be the quantity of matter in the Calcium Bromide

M is the atomic weight and m is the mass

M of CaBr2 is the sum of the atomic weight of its components

Mass of  Ca = 40.1 , Mass of Br = 79.9

M = 40.1 + 2×79.9

  0.422 = m/ (40.1+2×79.9)

  0.422 = m/ 199.9

  m = 0.422 × 199.9

  m = 84.35 g which is 88.4 g approximatively .

Thus ,88.4 g approximatively is  the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .

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The relative atomic mass of dihydrogen monoxide is 18g/mol

According to the periodic table:

Where; H = 1g/mol and O = 16g/mol

H2O = 1(2) + 16

= 2 + 16

= 18g/mol

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Answer:

“Magma” is exclusively found and formed beneath the earth’s surface. Once magma is on or above the surface of the earth it is referred to as “lava.” Magma is typically formed by extreme temperature melting solid rock within the earth. Pressure and rock composition can also affect magma formation. High pressure can help magma be “squeezed” from partially molten rock. Likewise, as rocks are usually composed of different minerals with different melting points, magma formation from rocks is usually only partial and uneven.

Explanation:

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

Answer:

a design is remodeled after analysis and tested again

Explanation:

Answer:

atoms contain tiny negatively charged subatomic particles or electrons

hope it will help you

.

.

.

hope it will helps

3 2 1 is the set of quantum numbers.

The set of numbers used to describe the position and energy of the electron in an atom is called quantum numbers. There are four quantum numbers, namely, principal, azimuthal, magnetic, and spin quantum numbers.

The rules for quantum numbers are: (n) can be any positive, nonzero integral value. (l) can be zero or any positive integer but not larger than (n-1). l = 0, 1, 2, 3, 4, …. (n-1) (ml) values follow the equation.

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Answer:

C

Explanation:

According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.

Answer:

I think it's C

Explanation:

Please, tell me if I'm incorrect.

→ Takli

→ Charkha

Answer:

Following are the answer to this question:

Explanation:

For the reductive amination of its carbonyl group, amino acids could be synthesized by reducing ammunition, which can be synthesized in the given attachment file:

please find the attachment:

Answer:

2, hydrogen

Explanation:

i think

Answer:

Answer is not hydrogen

Explanation:

did the test and got it wrong

Answer:

Explanation:

Fe2+ Has 4 unpaired electrons.

By method of elimination;

Option A: Ni2+ has two unpaired electrons. so this option is wrong.

Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.

Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.

Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.

Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.

Explanation:

the molecular weight of K2SO3 is 158. 2598 m/s.

Answer: A

Explanation: Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH, Urine pH= 4.90

Answer:

[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]

Explanation:

You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.

1. Moles of B₄H₁₀

[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]

2. Molecules of B₄H₁₀

[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]

3. Atoms of H

[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]

Answer:

OPTION C is correct

C) 1.07 g

Explanation:

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)

But we know molarity molarity= number of moles of solute/ volume of the solution

M= n/V

From the equation above

number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.

Then

1 mole of Cacl2 yields 2 moles of Agcl2

0.00375 moles of Cacl2 will produce let say Y.

Y= (0.00375 ×2)/1

= 0.0075 moles.

Number of moles of Agcl2 = mass /molar mass of Agcl

Molar mass of Agcl = 178

Then mass = 178 ×0.0075 = 1.047

Therefore, the mass of agcl precipitate is

1.07 g

Answer:

195.5 K

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2S + 3O2 + 2H2O → 2H2SO4

Next, we shall determine the number of mole in 17.55 g of H2SO4. This can be obtained as follow:

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98 g/mol

Mass of H2SO4 = 17.55 g

Mole of H2SO4 =...?

Mole = mass /Molar mass

Mole of H2SO4 = 17.55/98

Mole of H2SO4 = 0.179 mole.

Next, we shall determine the number of mole of O2 needed for the reaction. This can be obtained as illustrated below:

From the balanced equation above,

3 moles of O2 reacted to produce 2 moles of H2SO4.

Therefore, Xmol of O2 will react to produce 0.179 moleof H2SO4 i.e

Xmol of O2 = (3 x 0.179) / 2

Xmol of O2 = 0.2685 mole

Therefore, 0.2685 mole of O2 is needed for the reaction.

Finally, we shall determine the temperature of O2. This can be obtained by using the ideal gas equation as follow:

Volume (V) of O2 = 5.01 L

Pressure (P) = 0.860 atm

Number of mole (n) of O2 = 0.2685 mole

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =..?

PV =nRT

0.860 x 5.01 = 0.2685 x 0.0821 x T

Divide both side by 0.2685 x 0.0821

T = (0.860 x 5.01) /(0.2685 x 0.0821)

T = 195.5 K

Therefore, the temperature of O2 must be 195.5 K.

Answer:

- Weight 333.3 grams of iodine.

- Measure 500 mL of water.

- Vigorously mix the resulting solution.

Explanation:

Hello,

In this case, since 500 mL of a 40% (w/v) aqueous solution iodine is required, we can compute the required mass of iodine by defining the given mass-volume percent:

[tex]\% w/v=\frac{m_{iodine}}{m_{solution}} *100%=\frac{m_{iodine}}{m_{water}+m_{iodine}} *100%[/tex]

In such a way, we need to find mass of iodine, which is computed as:

[tex]m_{iodine}=\frac{\%w/v*m_{water}}{100w/v-\%} \\\\m_{iodine}=\frac{40*500}{100-40}\\ \\m_{iodine}=333.3g\\[/tex]

Thereby, the procedure will be:

- Weight 333.3 grams of iodine.

- Measure 500 mL of water.

- Vigorously mix the resulting solution.

Best regards.