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DedPeter [7]
3 years ago
11

Calculate the mass of iron oxide that contains a trillion iron atoms

Chemistry
1 answer:
4vir4ik [10]3 years ago
4 0
<h3>i think its help</h3><h3>iron ( III )</h3><h3>oxide ( Fe,O )</h3>
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Nataly_w [17]

What is [H+] given that the measured cell potential is -0.464 V and the anode reduction ... What half-reaction occurs at the cathode during the electrolysis of molten ... PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l); E° = 1.69 V .... For the cell Cu(s)|Cu2+||Ag+|Ag(s), the standard cell potential is 0.46 V. A cell ... hopw this helps

3 0
3 years ago
Read 2 more answers
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0
3 years ago
Why am I single? I dont know why tell meeeeeeeeeee
guapka [62]

Answer:

it's because your to fine

3 0
3 years ago
Explain how carbon is conserved in the carbon cycle.
Sonja [21]

Most of the carbon is put away in sedimentary carbonates and kerogens, with the rest being spread between the sea, the air, biomass, for example, plants and creatures, and petroleum products

<u>Explanation</u>:

  • The carbon cycle is the procedure where carbon goes from the surrounding into living beings and to the Earth and then again goes into the air. Plants take carbon dioxide from the air and use it for food preparation. Creatures at that point eat the nourishment and carbon is put away in their bodies or discharged as CO2 through the breath.  
  • Most of the carbon is put away in sedimentary carbonates and kerogens, with the rest being spread between the sea, the air, biomass, for example, plants and creatures, and petroleum products. This is known as carbon storage.  
  • For instance, carbon, a fundamental component in natural particles, is preserved as it is moved from inorganic carbon in a biological system to natural atoms in living life forms of the biological system and back as inorganic carbon to the earth.
6 0
4 years ago
When would formation of a derivative be a necessary step in this experiment?
const2013 [10]
<span>The formation of a derivative being a necessary step in the experiment lies in the importance of the derived structure. Often the derived product confers to reaction pathways which uses less reactive starting materials and more easily proceeds to completion. This also allows us to take a small amount of sample. The derived product at times is a general compound allowing its easy analysis. Often we encounter a product but we find it difficult to analyse it in ways we want. Here lies the essence of forming a derivative which often are simpler compounds allowing easier analysis yet having similar functional groups and structural properties. Also sometimes we encounter problems when our desired product is unstable and forms stable degraded products. But if we somehow manage to synthesize a derivative it may be relatively stable and form no degradation products. It would be stable at least for a significant period of time making it easier to study its properties. The derived product also at times are synthesized using general reaction pathways facilitating a way of easier synthesis and helping it to correlate with other similar reaction pathways and products.So the above paragraph accounts for the need of derivatives. When we encounter problems similar to those mentioned above it becomes necessary for a researcher to form rather synthesize a derivative.</span>
6 0
4 years ago
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