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GrogVix [38]
3 years ago
14

Cavendish’s name for hydrogen gas was "inflammable air." The word inflammable means "to burn. "Why did Cavendish use this name?

Chemistry
2 answers:
kozerog [31]3 years ago
6 0

The gas formed in the experiment could catch fire easily. So, he named the gas inflammable air.

bazaltina [42]3 years ago
3 0
He used that name because if you lit hydrogen on fire it would catch. therefore, it was dangerous to be around. think of it as a kind of warning. :) hope that helps.
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What is the approximate age of the oldest fossils of early human ancestors?
Vesna [10]

Answer:approx 300,000 years ago

Explanation:

Fossils attributed to H. sapiens, along with stone tools, dated to approximately 300,000 years ago, found at Jebel Irhoud, Morocco yield the earliest fossil evidence for anatomically modern Homo sapiens

7 0
3 years ago
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What is the molecular mass for a non-electrolyte if 35.0 g of it is dissolved in 45.0 grams of water and the solutions boiling p
adelina 88 [10]

The boiling point of water increases as the amount of impurities dissolved in it increases. For our purposes, we will consider the non-electrolyte to be the dissolved impurity. The change in the boiling point can be calculated using the equation:

\Delta T_b = i \times K_b \times m

where \Delta T_b is the change in boiling point, i is the van ‘t Hoff factor (whose value denotes the number of particles each formula unit of the dissolved substance dissociates into in water), K_b is the boiling point elevation constant, and m is the molality (moles of solute/kilogram of solvent) of the solution.

Right off the bat, since we're dealing with a non-electrolyte, the dissolved substance can be assumed not to dissociate in water. So, our van ‘t Hoff factor, i, would be 1 (by contrast, the i for an ionic compound like NaCl would be 2 since, in water, NaCl would dissociate into two particles: one Na⁺ ion and one Cl⁻ ion). We're also given our K_b, which is 0.51 °C/<em>m</em>.

Assuming the normal boiling point of pure water to be 100 °C (a defined value for sig fig purposes), the change in boiling point from having dissolved 35.0 g of the non-electrolyte can be obtained by subtracting 100 °C from the final—elevated—boiling point of 101.25 °C:

\Delta T_b = 101.25\text{ }^o\text{C} - 100\text{ }^o\text{C} = 1.25\text{ }^o\text{C}

Now, recall what we're asked to determine: the molecular mass of the dissolved substance. There is one unknown left in the equation: the molality of the solution. Let's first solve for that:

m = \frac{\Delta T_b}{K_b} = \frac{1.25^\text{ o}\text{C}}{0.51^\text{ o}\text{C}/m} \\ m = 2.45 \text{ mol solute/kg water}.

Notice that we didn't include the <em>i </em>since its value is 1.

Now, what would happen if we multiplied our molality by the mass of water we've been given? We would be left with the moles of solute. And what are we asked to find? The molecular mass, or the mass per mole. We can accomplish this in two steps. Remember to convert your mass of water to kilograms:

2.45 \text{ mol solute/kg water} \times 0.045 \text{ kg water} = 0.110 \text{ mol solute.}

And, finally, we divide the mass of our solute by the number of moles of solute:

\frac{35.0 \text{ g solute}}{0.110 \text{ mol solute}} = 317.5 \text{ g/mol}

Our answer to two significant figures (which is the number of sig figs to which our K_b is given) would be 320 g/mol.

5 0
3 years ago
"Elements that usually lose, or give away, electrons easily are ____"
Oxana [17]
B I took something similar before
3 0
2 years ago
In a series circuit, if a lamp breaks or a component is disconnected, the circuit is broken and all the components _________.
oksano4ka [1.4K]
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5 0
3 years ago
how would the equilibrium change if the temperature were increased in SO2(g)+NO2(g) --&gt; SO3(g) +heat?
kifflom [539]

Answer:

D. The amount of nitrogen dioxide and sulfur dioxide would increase.

Explanation:

Because the reaction is exothermic, heat is a product. According to Le Chaetelier's Principle, a system in equilibrium will adjust itself so as to relieve any stress placed upon it. If the temperature of the system increases, there are more products, and therefore the system is no longer at equilibrium. Thus, it will shift to favor the reactant side of the equation, where sulfur dioxide and nitrogen dioxide are located. This makes the answer D.

Edit: Brainly thinks that the correct spelling of 'Le Chaetelier' is a bad word so that's why it is spelled wrong lol.

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2 years ago
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