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BigorU [14]
2 years ago
10

The information below describes a redox reaction. 3 equations: First: upper C r superscript 3 plus (a q) plus 2 upper C l supers

cript minus (a q) right arrow upper C r (s) plus upper C l subscript 2 (g). Second: 2 upper C l superscript minus (a q) right arrow upper C l subscript 2 (g) plus 2 e superscript minus. Third: upper Cr superscript 3 plus (a q) plus 3 e superscript minus right arrow upper C r (s). What is the final, balanced equation for this reaction? 2 upper C r superscript 3 plus (a q) plus 6 upper C l superscript minus (a q) right arrow 2 upper C r (s) plus 3 upper C l subscript 2 (g). 2 upper C r superscript 3 plus (a q) plus 2 upper C l superscript minus (a q) plus 6 e superscript minus right arrow upper C l subscript 2 (g) plus 2 upper C r (s). Upper C r superscript 3 plus (a q) plus 6 upper C l superscript minus (a q) plus 3 e superscript minus right arrow 2 upper r (s) plus 3 upper C l subscript 2 (g). Upper C r superscript 3 plus (a q) plus 2 upper C l superscript minus (a q) right arrow upper C r (s) plus upper C l subscript 2 (g).
Chemistry
1 answer:
Marrrta [24]2 years ago
7 0

Answer:

A.

Explanation:

I just took the quiz.

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Mekhanik [1.2K]
Δt= \frac{NB_0 \pi a^2 }{IR}

This because ε = dΦ/dt, and emf is a voltage so it can also be written as dΦ/dt = IR

Because ΔΦ= ABcosθ for each of the N loops and cosθ is just 1 here only A needs to be replaced, and it can be replaced with A=πr^2, which makes that part of the equation ΔΦ_m=N*B0*πa^2
7 0
2 years ago
0.15 gm of metallic oxide was dissolved in 100 ml of 0.1 N H2SO4 and 25.8 ml of 0.095N NaOH were used to neutralise the remainin
Mila [183]

Answer:

Explanation:

25.8 ml of .095 N NaOH is needed to neutralise the remaining acid

equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .

acid remaining = .002451 gm equivalent .

acid initially taken = 100 ml of .1 N  /  1000  = . 01 gm equivalent

acid reacted with metal = .01  -.002451 = .007549 gm equivalent

This must have reacted with same gram equivalent of metal oxide

.007549  gm equivalent = .15 gm of metal oxide

1 gm equivalent = 19.87 gm

equivalent weight of metal = 19.87 - equivalent weight of oxygen

= 19.87 - 8 = 11.87 .

1

7 0
2 years ago
4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What
nikitadnepr [17]

<u>Answer:</u> The specific heat of metal is 2.34 J/g°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 4.82 g

m_2 = mass of water = 35 g

T_{final} = final temperature = 34.5°C

T_1 = initial temperature of metal = 115°C

T_2 = initial temperature of water = 28.7°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]

c_1=2.34J/g^oC

Hence, the specific heat of metal is 2.34 J/g°C

4 0
1 year ago
Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so
BartSMP [9]

Answer:

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Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

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If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

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  • NaOCl/HOCl (pKa = 7.49)
  • NaNO₂/HNO₂ (pKa = 3.35)
  • NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

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Light acts as a wave so when you burn a certain element it generates a specific wavelength which represents a specific color light. ^-^
7 0
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