Answer: The metalloids are a unique group of elements that share properties of both metals and nonmetals.
Explanation: They're also called the semimetals because of the shared properties of these elements along the dividing line between metals and nonmetals.
Metal rusts when it oxidized around moisture.
Answer:
2.47L
Explanation:
Using the combined gas law equation as follows:
P1V1/T1= P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 705mmHg
P2 = 760mmHg (STP)
V1 = 3.00L
V2 = ?
T1 = 35°C = 35 + 273 = 308K
T2 = 273K (STP)
Using P1V1/T1= P2V2/T2
705 × 3/308 = 760 × V2/273
2115/308 = 760V2/273
Cross multiply
308 × 760V2 = 2115 × 273
234,080V2 = 577,395
V2 = 577,395 ÷ 234,080
V2 = 2.47L
Answer:
Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:
Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:
The net equation is then:
However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:
Actual cathode half-equation:
Actual net reaction:
Answer:
See explanation.
Explanation:
Hello there!
In this case, considering the given information, it is not possible to figure out such order of reaction with respect to water as it is not involved in the given chemical reaction:
2E(g)+ F(g)+O(g)_
However, we can say that the order of reaction is 2 with respect to E (power of 2 in the rate law), 1 with respect to F (power of 1 in the rate law) and 0 with respect to O (power of 0 or not present in the rate law).
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