Identify the oxidizing and reducing agent in the following reaction, and determine which element is oxidized and which is reduced. Fe2+(aq) + ClO2(aq) → Fe3+(aq) + ClO2–(aq) ClO2 is the reducing agent and Fe2+ is the oxidizing agent . Fe is oxidized and Cl is reduced. g

Answer:

We know that there are 6.022 * 10²³ atoms in 1 mole of any element

So, that means that there are 6.022 * 10²³ atoms in 1 mole of Li (Lithium)

Atoms in 3.7 Moles:

atoms in one mole = 6.022 * 10²³

atoms in 3.7 moles = 6.022 * 10²³ * 3.7

atoms in 3.7 moles = 22.28 * 10²³

atoms in 3.7 moles = 2.228 * 10²⁴

Answer:

This question is incomplete

Explanation:

This question is incomplete but the answer to (1) and steps involved in calculating (2) and (3) will be discussed in the attachment

1. The reaction is an exothermic reaction because heat was given off. Exothermic reaction is a reaction in which heat is released from a reaction into the surroundings. And according to the question, the student observed that the temperature of the water increased during the course of the minutes.

Answer:

3.80 × 10²⁴ atoms of F

Explanation:

Step 1: Given data

Moles of barium fluoride (BaF₂): 3.16 mol

Step 2: Calculate the moles of fluorine in 3.16 moles of barium fluoride

The molar ratio of BaF₂ to F is 1:2. The moles of F in 3.16 moles of BaF₂ are 2/1 × 3.16 mol = 6.32 mol.

Step 3: Calculate the atoms of F in 6.32 moles of F

We will use Avogadro's number: there are 6.02 × 10²³ atoms of F in 1 mole of atoms of F.

6.32 mol × (6.02 × 10²³ atom/1 mol) = 3.80 × 10²⁴ atom

Answer:

7

Explanation:

Let x represent the number of moles of water in the hydrated salt i.e MgSO₄.xH₂O

The following data were obtained from the question:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mole of H₂O = x =?

Next, we shall determine the mass of water, H₂O in the hydrated salt, MgSO₄.xH₂O. This can be obtained as follow:

Mass of MgSO₄.xH₂O = 1.547 g

Mass of anhydrous MgSO₄ = 0.7554 g

Mass of H₂O =?

Mass of H₂O = (Mass of MgSO₄.xH₂O) – (Mass of anhydrous MgSO₄)

Mass of H₂O = 1.547 – 0.7554

Mass of H₂O = 0.7916 g

Finally, we shall determine the value of the x as illustrated below:

Mass of MgSO₄.xH₂O = 1.547 g

Molar mass of MgSO₄.xH₂O = 24 + 32 + (16×4) + x[(2×1) + 16]

= 24 + 32 + 64 + x(2 + 16)

= 120 + 18x

Mass of H₂O = 0.7916 g

Molar mass of xH₂O = 18x

Molar Mass of xH₂O/ Molar mass of MgSO₄.xH₂O = mass of xH₂O /Mass of MgSO₄.xH₂O

18x/ 120 + 18x = 0.7916/1.547

Cross multiply

0.7916 (120 + 18x) = 18x × 1.547

94.992 + 14.2488x = 27.846x

Collect like terms

94.992 = 27.846x – 14.2488x

94.992 = 13.5972x

Divide both side by 13.5972

x = 94.992 / 13.5972

x = 7

Thus, the formula for the hydrated salt, MgSO₄.xH₂O is MgSO₄.7H₂O

Number of moles of water, H₂O in the hydrated salt MgSO₄.7H₂O is 7.

The number of moles of attached water molecules is 7.

Mass of hydrated MgSO4 = 1.547 grams

Mass of anhydrous MgSO4 = 0.7554 grams

Number of moles of hydrated MgSO4 = 1.547 grams/120 + 18x

Number of moles of anhydrous MgSO4 = 0.7554 grams /120

Number of moles of anhydrous salt = Number of moles of hydrated salt

0.7554 grams /120 = 1.547 grams/120 + 18x

0.7554(120 + 18x) = 1.547 × 120

90.6 + 13.6x = 185.6

185.6 - 90.6 /13.6 = x

x = 7

The number of moles of attached water molecules is 7.

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Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\ [/tex]

volume = final volume of water - initial volume of water

volume = 60 - 25 = 35 mL

From the question we have

[tex]density = \frac{140}{35} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

Explanation:

In my opinion the answer should be SO2

Answer:

a should be answer i think.

Explanation:

from

de Broglie wavelength= h/ mv

substitute the values

joules= mass(kg) × acceleration (m/s^2) × distance(m)= kgm^2/s^2.

sorry I don't have my calc with me

According to De Broglie's Hypothesis, matter has both particle and wave properties. The wavelength of De Broglie is given as λ = h p,

wavelength (in m) of a proton =  11.09 x[tex]10^ {-55[/tex] m.

λ= h m v.

= 6.63 x [tex]10^{-34}[/tex] x  1.673 x [tex]10^{-24[/tex]x 1000

= 11.09 x[tex]10^ {-55[/tex] m

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Answer: Membrane Protiens

Explanation:

In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers. A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.

2x = 350- x

3x = 350

x = 350/x

therefore,

x = 116.67cm.

so, (350 - x)cm³ = (350- 166.67)cm³ = 233.33cm³

Hence distance covered by gas A is 116.67cm that by B is 233.33cm

Answer:

[a]. 55.5kmol/min.

[b]. The average velocity = 8.3m/s.

pipe diameter = 1.0954m.

Explanation:

So, from the question above we are given the following data or information or parameters which are going to help us in solving the above question;

=> ''A gas stream that contains 1.50 mole% CO2 flows through a pipeline''

=> ''Twenty (20.0) kilograms of CO2 per minute is injected into the line.''

=> '' . A sample of the gas is drawn from a point in the line 150 meters downstream of the injection point and found to contain 2.3 mole% CO2''

=> ''molar gas density is 0.123k mol/m3''

[a].The gas flow rate (k mol/min) upstream of the injection point can be calculated as follows.

- Determine the mass balance= j₁ +[ 20 /.123 × 44] = j₃. Thus, the flow gas rate is given as;

w₁ ×  1.50 mole% CO2 + 3.6954= 2.3 mole% CO2 × w₃.

-3.610 = - 0.008w₁.

w₁ = 451.3 m³/min.

Thus, the flow rate = 0.123 × 451.3 = 55.51 kmol/min.

[b]. The average velocity = 150/ 18 seconds = 8.3 m/s.

[c]. Pipe Diameter can be calculated as follows;

0.123  × 8.3  × [3.12  × (pipe diameter)² / 4] = 55.51/60.

[1.0209  × 3.12]/4  × [pipe diameter]² = 0.9252.

0.8014065  ×  [pipe diameter]² = 0.9252.

[pipe diameter]² = 0.9252/ .8014065.

[pipe diameter]² = 1.15443.

[pipe diameter] = √1.15443.

[pipe diameter] = 1.2 m² = 1.0954m.

A) The gas flow rate (k.mol/min) upstream of the injection point is;

55.51 k.mol/min

B) The average velocity and the diameter of the pipe are respectively;

8.33 m/s and 1.072 m

Due to the fact that there is no reaction or accumulation in the pipeline, we can say that the input of CO₂ is equal to it's output. Thus;

Upstream + Injection = Downstream

A) If we assume that x is the flow rate of gas (kmol/min), then there is the we can write the balance as:

0.015x + (20/44) = 0.023(x + 20/44)

Expanding gives us;

0.015x + 0.4545 = 0.023x + 0.01045

Solving for x gives;

x = 55.51 kmol/min

B) Formula for velocity is;

v = distance/time

Thus, since the length of the pipe is 150 m/s and time is 18 seconds, then velocity is;

v = 150/18

v = 8.33 m/s

Now, volumetric flow rate is calculated from;

Q = q' * ρ

where;

q' is gas flow rate

ρ is gas density

Thus;

Q = 55.51 * 0.123

Q = 451.3 m³/min = 7.522 m³/s

We can also express average velocity as;

v = Q/A

where A is area = πD²/4

Thus;

8.33 = 7.522/(πD²/4)

Making D the subject of the formula gives;

D = √((4 * 7.522)/(8.33π))

D = 1.072 m

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Answer:

What mass of sodium hydroxide is needed to react completely with 10.0 g of iron( III) chloride? The conversion sequence is: A. B.

Explanation:

make me as brain liest

Answer:

He, Cl, As, C, Na

Explanation:

^^

Answer:

1.53 L

Explanation:

Step 1: Given data

Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen

The molar mass of oxygen is 32.00 g/mol.

11.2 g × (1 mol/32.00 g) = 0.350 mol

Step 3: Calculate the volume of oxygen

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm

V = 1.53 L

Answer:

0.75

Explanation:

3/4=0.75 so therefore its 0.75

Answer:

Explanation:

Answer:

mountain, mountain range, mountain system, mountain belt

Explanation:

edge 2021

Answer:

371km

Explanation:

the lower the more fuel there is

Answer:

0.296j/g⁰c

Explanation:

we have the following information from this question before us.

mass iv substance = 50grams

we have initial temperature ti = 20.2⁰c

final temperature = 125⁰c

the energy that was provided = 155kj

we proceed with this formula

energy = mcΔT

1.55x10³ = 50 x c x (125-20.2)

1.55x10³ = c x 50gm x 104.8k

we divide through to get c

c = 1.55x10³/50g x 104.8

c = 0.296J/g⁰c

that is the specific heat of this substance.

thank you!

Answer:

The wallet is falling from a height of 1450 m

We are given that:

downward acceleration (a) = 10 m/s/s

downward displacement (s) = 1450 m

downward initial velocity (u) = 0 m/s (Since the wallet was dropped, it had no initial velocity)

horizontal initial velocity (v) = 13 m/s

Since there is no external force being applied on the wallet except the force of gravity, the wallet will keep moving at a constant velocity of 13 m/s

To calculate the horizontal distance travelled by the wallet, we need to know how long the wallet was airborne

To calculate how long the wallet was airborne, we have to find out that how long it took the wallet to hit the ground

Time taken for the wallet to hit the ground:

From the second equation of motion:

s = ut + 1/2 at² ------------(for vertical motion of the wallet)

1450 = (0)(t) + 1/2 (10)(t)²

1450 = 5t²

t² = 290

t = 17 (approx)

Time taken for the wallet to hit the ground = 17 seconds

Horizontal distance travelled by the wallet:

Since the wallet hits the ground after 17 seconds, it will move horizontally at a constant velocity of 13 m/s for 17 seconds

horizontal acceleration = 0 m/s/s

From the second equation of motion:

s = ut + 1/2at²

s = ut + 1/2 (0)t²

s = ut

here, u is the horizontal initial velocity of the wallet and the time taken by the wallet to hit the ground is 't'

s = (13)(17)

s = 221 m

Hence, the horizontal distance travelled by the wallet while falling from a height of 1450 m is 221 m

Answer:

[tex]T=2.78x10^3 \°C[/tex]

Explanation:

Hello,

In this case, considering that the safe temperature may be computed via the ideal gas law as we now the pressure, mass and volume via the dimensions:

[tex]V=\pi r^2 h=\pi *(41.0cm)^2*49.2cm=2.60x10^5cm^3*\frac{1L}{1000cm^3} =260L[/tex]

The pressure in atm is:

[tex]P=3.70MPa*\frac{1x10^6Pa}{1MPa} \frac{1atm}{101325Pa} =36.5atm[/tex]

And the moles considering the mass and molar mass (66 g/mol) of dinitrogen difluoride (N₂F₂):

[tex]n_{N_2F_2}=2.50kg*\frac{1000g}{1kg}*\frac{1mol}{66g} =37.9mol[/tex]

In sich a way, by applying the ideal gas equation, which is not the best assumption but could work as an approximation due to the high temperature, the temperature, with three significant figures, will be:

[tex]T=\frac{PV}{nR}=\frac{36.5Pa*260L}{37.9mol*0.082\frac{atm*L}{mol*K} }\\ \\T=3053.6K-273.15\\\\T=2.78x10^3 \°C[/tex]

Best regards.

Answer:

corbon 12

Explanation:

carbon atoms with the usual 6 nutrons have a mass number of 12

6 protons plus 7 neutrons equals 13 for carbon nuclei with 7 neutrons, these atoms make up the carbon-13 element.

A naturally occurring stable isotope of carbon with a nucleus comprising six protons and seven neutrons is carbon-13 (13C). It makes up roughly 1.1% of all the native carbon on Earth and is one of the environmental isotopes.

Since they are carbon atoms with extra neutrons, carbon 12, 13, and 14: You could also say that the difference in mass between these isotopes is molecular mass.

Therefore, molecular mass is determined by adding the number of protons and neutrons (if you have 1 mole of each isotope, carbon 14 would have the greatest mass).

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Answer:

Specific heat of aluminum  = 0.27 j/g.°C

Explanation:

Given data:

Mass of Al = 100 g

Initial temperature = 108.4°C

Final temperature = 68.2°C

Heat released = -1080 j

Specific heat of aluminum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  68.2°C -  108.4°C

ΔT = - 40.2°C

Q = m.c. ΔT

-1080 j = 100 g ×c ×- 40.2°C

-1080 j = -4020 g.°C ×c

c = -1080 j/-4020 g.°C

c = 0.27 j/g.°C

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water  =  0.28 moles   (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)

Answer:

[tex]Ka_1=1.61x10^{-3}[/tex]

[tex]Ka_2=1.01x10^{-8}[/tex]

Explanation:

Hello.

In this case, since the stepwise dissociation of malonic acid which is a diprotic acid that we are going to symbolize by H₂A, is:

[tex]H_2A\rightleftharpoons H^++HA^-;Ka_1\\\\HA^-\rightleftharpoons H^++A^-;Ka_2[/tex]

The first ionization has the following equilibrium expression:

[tex]Ka_1=\frac{[H^+][HA^-]}{[H_2A]}[/tex]

Whereas the concentration of H⁺ equals the concentration of HA⁻ and is computed via the pH:

[tex][H^+]=[HA^-]=10^{-pH}=10^{-1.47}=0.0339M[/tex]

Next, we compute the molarity of the 19.5 g of malonic acid (molar mass = 104.06 g/mol) as shown below:

[tex][H_2A]=\frac{19.5g/(104.06 g/mol)}{0.250L}=0.750M[/tex]

Thus, Ka1 turns out:

[tex]Ka_1=\frac{(0.0339)(0.0339)}{0.750-0.0339}=1.61x10^{-3}[/tex]

Now, for the second ionization, since the 0.300-M sodium hydrogen malonate is the source of HA⁻, and the pH is 4.26, we can compute the concentration of both H⁺ and A⁻² again by considering the pH:

[tex][H^+]=[A^-^2]=10^{-4.26}=5.50x10^{-5}M[/tex]

Therefore Ka2 turns out:

[tex]Ka_2=\frac{[H^+][A^{-2}]}{[HA^-]}=\frac{(5.50x10^{-5})(5.50x10^{-5})}{0.300-(5.50x10^{-5})}\\ \\Ka_2=1.01x10^{-8}[/tex]

Best regards!

Answer:

Required heat = 2,259.36 N (approx)

Explanation:

Given:

Mass = 135 g

Specific heat (c) =4.184 J/gC

Change in temperature ΔT = 27-23 = 4 c

Find:

Required heat

Computation:

Q = mcΔT

Q = (135)(4.184)(4)

Required heat = 2,259.36 N (approx)

The heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J

To solve this question, we'll begin by calculating the change in the temperature of the reservoir. This can be obtained as follow:

Initial temperature (T₁) = 23 °C

Final temperature (T₂) = 27 °C

ΔT = T₂ – T₁

ΔT = 27 – 23

Finally, we shall determine the heat transferred to the reservoir.

Change in temperature (ΔT) = 4 °C

Mass (M) = 135 g

Specific heat capacity (C) = 4.184 J/gºC

Q = MCΔT

Q = 135 × 4.184 × 4

Therefore, the heat transferred by the chemical reaction to the reservoir of a calorimeter containing 135 g of dilute aqueous solution is 2259.36 J

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Answer:

LA PRIMERA ES N

Explanation:

Answer:

The solution to this question can be defined as follows:

Explanation: