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2 years ago

Which physical property can be measured??

Chemistry

2 answers:

Scorpion4ik [409]2 years ago

5 0

Answer:

Density for Edge

Explanation:

Density is the only choice that has a numerical answer.

Nikitich [7]2 years ago

4 0

Extensive property (like volume and mass), can be measured.

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Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?

konstantin123 [22]

Answer: Ammonia (NH3) and sodium carbonate (Na2CO3), because they accept hydrogen ions but lack hydroxide ions.

Explanation:

i took the test and got it correct :) hope this helps

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2 years ago

How do you calculate the relative formula mass of a compound

RSB [31]

Write the formula of the compound.
Write the numbers of each atom in the formula. Insert the relative atomic mass for each type of atom. Calculate the total mass for each element.
Add up the total mass for the compound.

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2 years ago

Mammals use their lungs to provide oxygen for themselves from the air. Fish spend a majority of their time in the water, and ins

Elenna [48]

B of course is the answer

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2 years ago

Read 2 more answers

What mass of O2, in grams, is required to completely react with 0.025 g C3H8?

8090 [49]

2C3H8+ 702--->6CO2+8H20
FROM Equation above 2 moles of C3H8 reacted with 7 moles of oxygen to form 6 moles of c02 plus 8 molesof H2O
the moles of c3H8 reacted is = MASS/ R.F.M
THE R.F.M =48+8=44
Number of moles is hence 0.025/44=5.68x10^-4
since ratio of C3H8 to O2 is 2:7 Therefore moles of O2 reacted is 1.989 x10^-3
mass= r.f.m x number of moles
(1.989x10^-3) x 32 =0.064g

6 0

2 years ago

Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends

IgorLugansk [536]

24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its kinetic energy.

$E_\text{k} = 1/2 \; m \cdot v^{2}$

Where

$E_\text{k}$ is the kinetic energy of the molecule,
$m$ its mass, and
$v^{2}$ the square of its speed.

Besides, the kinetic theory of gases suggests that for an ideal gas,

$E_\text{k} \propto T$

where $\text{T}$ its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the average kinetic energy of molecules shall be the same for any ideal gas at the same temperature. So is the case for HCl and NH₃

$E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)$

$m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

Where

$m(\text{HCl})$ , $v(\text{HCl})$ , and $E_\text{k}(\text{NH_3})$ the mass, speed, and kinetic energy of an HCl molecule;
$m(\text{NH}_3)$ , $v(\text{NH}_3)$ , and $E_\text{k}(\text{NH}_3)$ the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their molar mass. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, $m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3)$ . Therefore:

$36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

$36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)$

$\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)$

The average speed NH₃ molecules would be $\sqrt{36.45/17.03} \approx 1.463$ if the average speed of HCl molecules $v(\text{HCl})$ is 1.

$\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}$

$\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}$

8 0

2 years ago