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kupik [55]
3 years ago
15

Consider the following reaction of calcium hydride (CaH2) with molten sodium metal: CaH2(s) + 2 Na(l) -> 2 NaH(s) + Ca(l) Ide

ntify the species being oxidized and the species being reduced?
Chemistry
1 answer:
vodka [1.7K]3 years ago
4 0

Answer:

This is not a redox reaction. None of the species are reduced and none are oxidized

Explanation:

In a redox reaction at least one species must be oxidized and another reduced. You see this by a change in oxidation number. In this question the oxidation numbers are the same before and after the reaction.

Ca  2

H  -1

Na +1

I -1

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Most of the minerals in the Earth’s crust are a type of what mineral? A. mica B. gem C. metal D. silicate
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3 years ago
How much heat is required to heat 1.6g of ice from -16c to steam at 112c?
oksian1 [2.3K]

Answer:

Total heat ≅ 49.07 kJ

Explanation:

Given that:

mass = 1.6 g = 0.016 kg

Initial temperature = - 16 ° C

final temperature = 112° C

specific heat for ice = 2.06 kJ/kgC

specific heat of water = 4.186 kJ/kgC

heat fusion of ice = 334 kJ/kg

specific heat for steam = 2.1 kJ/kgK

heat of vaporization of water = 2256 kJ/kg

To heat ice from -16 ° C to 0 ° C

Q₁ =  2.06 kJ/kgC × 0.016 kg ×  16 ° C

Q₁ =  0.52736 kJ

To melt Ice at 0° C

Q₂= 334 kJ/kg × 0.016 kg = 5.344 kJ

To heat water from 0° C to  100° C

Q₃ = 4.186 kJ/kgC × 0.016 kg  × 100° C

Q₃ = 6.6976 kJ

To vaporize water to steam at 100° C

Q₄ = 2256 kJ/kg × 0.016 kg = 36.096 kJ

To heat steam from 100C to 112° C

Q₅ = 2.1 kJ/kgC × 0.016 kg × 12 C

Q₅ = 0.4032 kJ

Total heat = Q₁ + Q₂ + Q₃ + Q₄  + Q₅

Total heat =  (0.52736 +  5.344 +  6.6976 + 36.096 + 0.4032) kJ

Total heat = 49.06816  kJ

Total heat ≅ 49.07 kJ

6 0
3 years ago
ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n
ArbitrLikvidat [17]

Answer:

\boxed{\text{3.3 mL}}

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}

2. Moles of NaOH  

\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}

3. Molar concentration of NaOH

c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

c_{1}V_{1} = c_{2}V_{2}

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1   mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}

(b) Calculate the volume  of dilute solution

\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}

\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}

4 0
3 years ago
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