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soldi70 [24.7K]
3 years ago
6

How is 1 hectoliter different from 1 kiloliter?

Chemistry
2 answers:
LUCKY_DIMON [66]3 years ago
5 0
1 hectoliter is 26.4172

1 kiloliter is 264.172
tamaranim1 [39]3 years ago
3 0

Answer:

A hectoliter is 10 times smaller than a kiloliter

Explanation:

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vekshin1

Answer:

The sum of the molar masses of each isotope of the element.

7 0
3 years ago
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Which reaction is single replacement?
Tatiana [17]
<span>Among the given choices, the third option is the only one which illustrates single replacement. (3)H2SO4 + Mg --> H2 + MgSO4 A single replacement is also termed as single-displacement reaction, a reaction by which an element in a compound, displaces another element. It can be illustrated this way: X + Y-Z → X-Z + Y</span>
5 0
3 years ago
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How any moles of H2O will be produced from 12.3 moles of HCL reacting with Ca(OH)2?
stiks02 [169]

Answer:

\large \boxed{\text{12.3 mol HCl}}  

Explanation:

We need a balanced chemical equation with moles.

            2HCl +Ca(OH)₂ ⟶ CaCl₂ + 2H₂O

n/mol:    12.3

The molar ratio is 2 mol H₂O:2 mol HCl.

\text{Moles of H$_{2}$O} = \text{12.3 mol HCl} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{2 mol HCl}} = \textbf{12.3 mol HCl}\\\\\text{The reaction produces $\large \boxed{\textbf{12.3 mol HCl}}$}

3 0
3 years ago
3. What is the energy of a photon whose frequency is 5.2 x 1015 Hz? Use the equation: E = hxv
anzhelika [568]

Answer:

3. 3.45×10¯¹⁸ J.

4. 1.25×10¹⁵ Hz.

Explanation:

3. Determination of the energy of the photon.

Frequency (v) = 5.2×10¹⁵ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained by using the following formula:

E = hv

E = 6.626×10¯³⁴ × 5.2×10¹⁵

E = 3.45×10¯¹⁸ J

Thus, the energy of the photon is 3.45×10¯¹⁸ J

4. Determination of the frequency of the radiation.

Wavelength (λ) = 2.4×10¯⁵ cm

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm

2.4×10¯⁵ cm = 2.4×10¯⁷ m

Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m

Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:

Wavelength (λ) = 2.4×10¯⁷ m

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

v = c / λ

v = 3×10⁸ / 2.4×10¯⁷

v = 1.25×10¹⁵ Hz

Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.

8 0
3 years ago
Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p
Licemer1 [7]
                                                   N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g 
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4

</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react  0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
moles= \frac{grams}{molar mass}             (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0
3 years ago
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