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2 years ago

How many particles are in 5.0 ml of Fe

Chemistry

1 answer:

nordsb [41]2 years ago

7 0

f it is solid iron, use the density to get the mass. D x V = mass

Use the mass to find the moles of Fe. Mass/atomic mass = moles

Moles x Avogadro's number will give you the particles, assuming that you mean atoms.

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QUICK CHECK

Free_Kalibri [48]

Answer:

Tetraedral: CH4

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Trigonal planar: BF3

Trigonal pyramidal: NH3

Linear: CO2

5 0

2 years ago

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Select the correct answer.

sergejj [24]

D.) black text on a yellow background. There is much more contrast in the colors black and yellow than the other choices.

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2 years ago

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Calculate the mass of 2.50 mol of CH,OH(1). Show your work. Use the appropriate

Sophie [7]

Answer:

80.1 grams

Explanation:

Find the molar mass of CH3OH first by using the periodic table values.

12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H

=32.042 so that is the molar mass

Now that you have 2.50 moles of CH3OH, you can calculate the mass in g

2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105

32.042g / 1 mol is the same as 32.042 g/mol

Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH

7 0

2 years ago

7. Propane, C₂H₁, has become a popular gas to use in fireplaces.

kramer

Answer:

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8 0

1 year ago

The solubility of BaCO3(s) in water at a certain temperature is 4.4 10–5 mol/L. Calculate the value of Ksp for BaCO3(s) at this

azamat

Answer : The value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

Solution : Given,

Solubility of $BaCO_3$ in water = $4.4\times 10^{-5}mole/L$

The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.

The equilibrium equation is,

$BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3$

Initially - 0 0

At equilibrium - s s

The Solubility product will be equal to,

$K_{sp}=[Ba^{2+}][CO^{2-}_3]$

$K_{sp}=s\times s=s^2$

$[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L$

Now put all the given values in this expression, we get the value of solubility constant.

$K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2$

Therefore, the value of $K_{sp}$ for $BaCO_3$ is $19.36\times 10^{-10}mole^2/L^2$ .

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3 years ago