Answer:
B
Explanation:
Use your periodic table to find out the charge on a Fl ion if your periodic table is constructed that way. In any event, the charge on Fl when it becomes an ion is -1. I don't think it has any other charge. So all of the elements in column 17 will have at least -1 in common.
Group 17 elements, also known as halogens, have seven valence electrons and tend to gain one electron to become stable, forming a -1 ion.
The elements in Group 17 on the periodic table are also known as halogens. These elements, which include fluorine, chlorine, bromine, iodine, and astatine, have seven valence electrons. Hence, they tend to gain one electron to achieve a stable electron configuration, forming a -1 ion. Therefore, the correct answer to your question is B. -1 ion.
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Number of moles is defined as the ratio of given mass in grams to the molar mass of compound.
Number of moles =
Now, put the value of given mass of in grams and molar mass of in g/mol i.e. 13 g/mol.
Thus,
number of moles =
= 0.692 mol
Hence, number of moles of = 0.692 mol
Answer:
The specific heat of the alloy is 2.324 J/g°C
Explanation:
Step 1: Data given
Mass of water = 0.3 kg = 300 grams
Temperature of water = 20°C
Mass of alloy = 0.090 kg
Initial temperature of alloy = 55 °C
The final temperature = 25°C
The specific heat of water = 4.184 J/g°C
Step 2: Calculate the specific heat of alloy
Qlost = -Qwater
Qmetal = -Qwater
Q = m*c*ΔT
m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)
⇒ mass of alloy = 90 grams
⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED
⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C
⇒ mass of water = 300 grams
⇒ c(water) = the specific heat of water = 4.184 J/g°C
⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C
90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C
c(alloy) = 2.324 J/g°C
The specific heat of the alloy is 2.324 J/g°C
Using only the information above, can you calculate the solubility of X in water at 15°C ? If you said yes, calculate it.
Answer:
The answer is YES
The value is
Explanation:
From the question we are told that
The volume of the sample taken is v = 13.0 mL
The temperature is
The mass of the sample is
Generally the solubility of the substance X is mathematically represented as
=>
=>
=>
Answer:
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
Explanation:
Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.
In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
In a solution it is fulfilled:
Ci* Vi = Cf* Vf
where:
In this case:
Replacing:
1.40 M* 20 mL= 0.088 M* Vf
Solving:
Vf= 318.18 mL
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.
To dilute a solution, you can use the formula:
M1V1 = M2V2
where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. Rearranging the formula, we can solve for V2:
V2 = (M1 · V1) / M2
Plugging in the values given:
V2 = (1.40 M · 20.0 mL) / 0.0880 M = 318.18 mL
To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.
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Answer:
combustion reaction
Explanation:
there is oxides in the equation c3h8(g)+5o2(g)3co2(g)+4h2o(g)
Answer:
2.11 molecules of CO₂
Explanation:
Given data:
Mass of CO₂ = 15.2 g
Molecules of CO₂ = ?
Solution:
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 15.2 g/ 44 g/mol
Number of moles = 0.35 mol
Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
0.35 mol × 6.022 × 10²³ molecules / 1 mol
2.11 molecules of CO₂