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3 years ago

Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?

1 answer:

5 0

Mechanical twinning occurs in metals having body center cubic and hexagonal closed packed structures. Twinning is said to occur when a portion of a crystal takes up an orientation that is related to the orientation of the untwinned lattice in a definite symmetrical manner.

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An electrode has a potential of 1.201 V with respect to a saturated silver-silver chloride electrode. What would the electrodes

fredd [130]

Answer:

The potential wrt. calomel is 1.254 V

Explanation:

Given:

Potential wrt. silver chloride $E_{Ag} = 1.201$ V

Potential wrt. saturated silver chloride $E = 0.197$ V

Potential wrt. SCE $E_{Hg} = 0.241$ V

Now potential wrt. hydrogen is given by,

$= 1.201- 0.197$

$= 1.004$ V

And we find for potential wrt. calomel,

$=$ potential wrt. hydrogen + potential wrt. SEC

$= 1.004 +0.241$

$= 1.254$ V

Therefore, the potential wrt. calomel is 1.254 V

7 0

3 years ago

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCI. What volume of 3.00 M HCI

frozen [14]

Answer:

30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.

Explanation:

M1 = 3.00 M

M2 = 1.80 M

V2 = 50 .0 mL = 50 /1000 L = 0.05 L

V1 = unknown

In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.

So we use the equation:

M1 V1 = M2 V2

V1 = M2 V2 / M1

V2 = 1.80 * 0.05 / 3.0

V2 = 0.09 /3.0

V2 = 0.03 L or 30 mL

To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.

5 0

3 years ago

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

SpyIntel [72]

The empirical formula of compounds formed from the given ions are as follows:

Pb⁴⁺ = PbO₂
NH₄⁺ = NH₄Cl
CrO₄²⁻ = Na₂CrO₄
SO₄²⁻ = K₂SO₄

<h3>What is the empirical formula of a compound?</h3>

The empirical formula of a compound is the simplest formula of the compound showing the simplest ratios in which elements in the compound combine.

The empirical formula of compounds formed from the given ions are as follows:

Pb⁴⁺ = PbO₂
NH₄⁺ = NH₄Cl
CrO₄²⁻ = Na₂CrO₄
SO₄²⁻ = K₂SO₄

In conclusion, the empirical formula is the simplest formula of a compound.

Learn more about empirical formula at: brainly.com/question/1581269

#SPJ1

4 0

1 year ago

Helppppp pleaseeee xxxxxx

kogti [31]

Answer:

2812.6 g of H₂SO₄

Explanation:

From the question given above, the following data were obtained:

Mole of H₂SO₄ = 28.7 moles

Mass of H₂SO₄ =?

Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:

Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)

= 2 + 32 + 64

= 98 g/mol

Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:

Mole of H₂SO₄ = 28.7 moles

Molar mass of H₂SO₄ =

Mass of H₂SO₄ =?

Mole = mass / Molar mass

28.7 = Mass of H₂SO₄ / 98

Cross multiply

Mass of H₂SO₄ = 28.7 × 98

Mass of H₂SO₄ = 2812.6 g

Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄

3 0

3 years ago

What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin

Contact [7]

<u>Answer:</u> The total volume of the gaseous products is 1044.29 L

<u>Explanation:</u>

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = $\frac{1}{22.4}\times 116=5.18mol$ of butane

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

<u>For carbon dioxide:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = $\frac{8}{2}\times 5.18=20.72mol$ of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

<u>For water vapor:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = $\frac{10}{2}\times 5.18=25.9mol$ of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

3 0

2 years ago