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Maslowich
3 years ago
12

What are some different ways large or small distance can be measured?

Chemistry
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

The method used for measuring the small distance is by using the scales and the distance measured over long distance is by inch tape or measuring tape.

Explanation:

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Why do the metals Co, Rh, and Lr form octahedral complexes (rather than tetrahedral or square planar complexes)? Hint: Look at t
algol13

Explanation:

Octahedral complexes will be favoured over tetrahedral ones because:

It is more favourable to form six bonds rather than four

The crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.

The transition metals Co, Rh, Lr are in group 9 of d block and they have 3d, 4d, and 5d orbitals respectively

6 0
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Inessa [10]

Answer:

methane + oxygen → carbon dioxide + water

Explanation:

methane + oxygen → carbon dioxide + water

5 0
3 years ago
For question 9 I need help push or pull
adelina 88 [10]

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a push

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5 0
3 years ago
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
The volume of a gas-filled balloon is 30.0 L at 313 K and 1520 mm Hg pressure. What would the volume be at standard temperature
pashok25 [27]
313 k
step-by-step-explanation
4 0
3 years ago
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