Iron is a type of element because iron is metal and metal is also a mineral
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Answer:The answer is D.1,3
Explanation:
Answer:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
Explanation:
Hello,
In this case, the undergoing chemical reaction is:

Thus, the rate is given as:
![rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}](https://tex.z-dn.net/?f=rate%3D-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%20%5BHBr%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B%5CDelta%20%5BBr_2%5D%7D%7B%5CDelta%20t%7D%20%3D%5Cfrac%7B%5CDelta%20%5BH_2%5D%7D%7B%5CDelta%20t%7D)
It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.
Therefore, the answer is:
c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt
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The molecular structure of 1-nitrobutane is
. The structure of 1-nitrobutane is shown below.
An atom's formal charge would be determined by the covalent model of chemical bonding, which assumes that almost all chemical bonds include equal sharing of electrons among all atoms, regardless their relative electronegativity.
The structure for 1-nitrobutane, making sure to add all non-zero formal charges
There are four kind of molecule present in 1-nitrobutane and they are carbon, hydrogen , nitrogen and oxygen. Nitrogen is bonded with two oxygen atom out of them one oxygen atom is attached with single bond and second oxygen atom is bonded with double bond. Nitrogen has positive charge whereas oxygen has negative charge.
It is a kind of alkane in with nitro group is attached with alkane group.
To know more about 1-nitrobutane
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You can establish a system of two equation with two variables.
Varibles are:
V1 = volume of the 50% sugar solution
V2 = volumen of the 80% sugar solution
Equations:
Balance of sugar:
Sugar from 50% solution: 0.5*V1
Sugar from 80% solution: 0.8*V2
Sugar in the final solution (mix): 0.6 * 105 = 63
1) 0.5V1 + 0.8V2 = 63
Final volume = volume of 50% solution + volume of 80% solution
2) V1 + V2 = 105
From (2) V1 = 105 - V2
Substitue in (1)
0.5 (105 - V2) + 0.8 V2 = 63
52.5 - 0.5V2 + 0.8V2 = 63
0.3 V2 = 63 - 52.5
0.3 V2 = 10.5
V2 = 10.5/0.3
V2 = 35mL
V1 = 105 - 35 = 70 mL
Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.