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BabaBlast [244]
3 years ago
13

Which temperature change would cause a sample of an ideal gas to double in volume while pressure is held constant? A. From 400K

to 200K B. From 200K to 400K C. From 400 C to 200C D. From 200C to 400C
Chemistry
1 answer:
Advocard [28]3 years ago
7 0

Answer:

The correct answer is option B. From 200K to 400K

Explanation:

Hello!

Let's solve this!

The ideal gas equation is the following PV = nRT

Where the variables are:

P: pressure

T: temperature

V: volume

R: universal constant of ideal gases

n: number of moles

We see in the equation the proportional relationship of volume with temperature.

We deduce that if the volume increases twice, then the temperature increases twice.

The temperature has to be in Kelvin, that's why we chose that answer.

Of the options, the correct answer is option B. From 200K to 400K

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If an atom has three shell and number of valence electrons are two. Is this element metal
cluponka [151]

Answer:

Metal

Element: Calcium

Valency: 2

Explanation:

To find the element, let's use the periodic table. (Look below)

We already went past 3 shells, just need the 2 electrons after it.

Just skip to the 4th row and count 2 to the right

We end up at Calcium.

Calcium is a metal and we're also on the alkaline earth metals column.

Calcium will need to lose 2 electrons to reach stability, so the valency is 2.

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3 years ago
Số oxi hoá của nitơ trong NH3, H2SO4, NO3 lần lượt là:
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Answer:

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3 years ago
One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2. The
sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

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3 years ago
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Answer:

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