Which of the elements shown will not form ions, and why will they not do so?

If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?

Y_Kistochka [10]

Answer:

-26.125 kj

Explanation:

Given data:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

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 = T2 - T1

ΔT = 5.0°C - 30.0°C

ΔT = -25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m.c. ΔT

Q = 250.0 g × 4.18 j/g.°C × -25°C

Q = -26125 j

J to kJ

-26125 j ×1 kj /1000 j

-26.125 kj

5 0

3 years ago

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

likoan [24]

Balanced Half reactions are:

At anode 2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$

At Cathode: 2 $H^{+}$ + $2e^{-}$ ==> H₂

Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, $H^{+}$ , $Mn^{2+}$ and $Cl^{-}$

Now at Anode reaction will occur as given:

2 $Cl^{-}$ ==> Cl₂+ $2e^{-}$ + H₂O ==> $ClO^{-}$ + 2 $H^{+}$ + $Cl^{-}$ (will occur)

At Cathode:

2 $H^{+}$ + $2e^{-}$ ==> H₂ (will occur)

At Cathode:

$Mn^{2+}$ + $2e^{-}$ ==> Mn (This reaction will not occur)

The deposition of solid Mn will not occur because in aqueous solution, $H^{+}$ will be reduced before $Mn^{2+}$ .

The reduction potentials for $H^{+}$ is zero whereas reduction potential for $Mn^{2+}$ is - 1.18V.

The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.

To learn more about the half reaction please click on the link brainly.com/question/13186640

#SPJ4

8 0

1 year ago

Use the following equation to answer question 1-4. Make sure you balance first.

worty [1.4K]

<h3>Answer:</h3>

5.2 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O

[Given] 10.4 mol HCl

<u>Step 2: Identify Conversions</u>

[RxN] 6 mol HCl = 3 mol H₂O

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 10.4 \ mol \ HCl(\frac{3 \ mol \ H_2O}{6 \ mol \ HCl})$
Multiply/Divide: $\displaystyle 5.2 \ mol \ H_2O$

4 0

3 years ago

I’LL MAKE YOU BRAINLIEST+ FREE POINTS

Pavel [41]

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

Ci*Vi=Cf*Vf

where C and V are concentration and volume, respectively; and the i and f subscripts indicate initial and final respectively.

In this case, being:

Ci= 7 M
Vi= 0.60 L
Cf= 2.5 M
Vf=?

Replacing:

7 M*0.60 L= 2.5 M* Vf

Solving:

$Vf=\frac{7 M*0.60 L}{2.5 M}$

Vf= 1.68 L

<u><em>The solution's new volume is 1.68 L</em></u>

4 0

3 years ago

20 points! Help please

Olin [163]

Answer: 2N2O3 -> 2N2 + 3O2

Explanation:

5 0

2 years ago

Read 2 more answers