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Leya [2.2K]
2 years ago
14

Two or more elements or compounds combine to make one complex substance

Chemistry
1 answer:
Leno4ka [110]2 years ago
4 0

Answer:

Synthesis reaction

Explanation:

Two or more elements or compounds combining to make one complex substance occurs in a synthesis reaction.

A synthesis or combination reaction is one that involves the formation of a single product from two or more reactants.

   A + B + C →  D

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Fe + H2SO4 = Fe2(SO4)3 + H2
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Which element is similar to the properties bromine?
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Answer: Fluorine

Explanation: It belongs in the same group as Bromine

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Suppose you want to create a 6 ng/μL solution in a 25 mL volumetric flask. However, this concentration cannot really be accurate
Over [174]

Answer:

Mass of chemical = 1.5 mg

Explanation:

Step 1: First calculate the concentration of the stock solution required to make the final solution.

Using C1V1 = C2V2

C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution

C1 = C2V2/V1

C1 = (6 * 25)/ 0.1

C1 = 1500 ng/μL = 1.5 μg/μL

Step 2: Mass of chemical added:

Mass of sample = concentration * volume

Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL

Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg

Therefore, mass of sample = 1.5 mg

4 0
2 years ago
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dmitriy555 [2]

Answer:

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3 0
3 years ago
3.47 g of the hydrated "double salt", ammonium iron (II) sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O was dissolved in 200. mL of wa
Ostrovityanka [42]

Answer:

1.4 × 10^-4 M

Explanation:

The balanced redox reaction equation is shown below;

5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O

Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol

Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles

Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M

Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M

Volume of Fe^2+ (VA)= 20.0 ml

Let the concentration of MnO4^- be CB (the unknown)

Volume of the MnO4^- (VB) = 12.6 ml

Let the number of moles of Fe^2+ be NA= 5 moles

Let the number of moles of MnO4^- be NB = 1 mole

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5

CB = 1.4 × 10^-4 M

7 0
3 years ago
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