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1 year ago

How many moles of Na+ ions are in 100.mL of 0.100M Na3PO4(aq)?

Chemistry

1 answer:

expeople1 [14]1 year ago

8 0

Answer:

The answer is C) 0.0300 mol

Explanation:

First you need to get the total amount of moles is dissolved in the solution. This can be obtained doing the following:

Molarity = moles of solute / liters of solution

moles of solute = molarity x liters of solution

First, the volume has to be in liters, then :

100. ml x (1 L / 1000 mL) = 0.100 L

(We then substitute)

moles of solute = 0.100 mol/L x 0.1 L = 0.0100 mol of solute (In this case Na3PO4)

Having the moles of the solute, we now need to find how many moles of Na+ ions are there.

We need the conversion factor of 3 Na+ moles per 1 mole of Na3PO4

We then find the amount of moles doing as follows:

0.0100 mol Na3PO4 x (3 mol Na+ / 1 mol Na3PO4) = <u>0.0300 mol Na+.</u>

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What is the mass of a sample of water containing 3.55×1022 molecules of h2o?

ad-work [718]

6,02×10²³ -------------- 18g
3,55×10²² -------------- x $x=\frac{3,55*10^{22}*18g}{6,02*10^{23}}=10,61*10^{22-23}g=10,61*10^{-1}g=1,061g$

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2 years ago

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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2 years ago

In 100 grams of sweet peas there are 14.5 carbohydrates,5.7 grams of sugars,5.1 grams of dietary fiber, 5.4 grams of protein and

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14.5 % carb
5.7% sugar
5.1% fiber
5.4% protein
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2 years ago

The weight percent of concentrated HClO4(aq) is 70.5% and its density is 1.67 g/mL. What is the molarity of concentrated HClO4

ollegr [7]

Answer:

[HClO₄] = 11.7M

Explanation:

First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.

Let's convert the mass to moles → 70.5 g . 1mol/100.45 g = 0.702 moles

Now we can apply the density to calculate the volume.

Density always refers to solution → Solution density = Solution mass / Solution volume

1.67 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.67 g/mL → 59.8 mL

To determine molarity (mol/L) we must convert the mL to L

59.8 mL . 1L/1000mL = 0.0598 L

Molarity → Moles of solute in 1L of solution → 0.702 mol / 0.0598 L = 11.7M

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2 years ago

Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.

icang [17]

Answer : $3.2 X 10^{15} g/cm^{3}$

Explanation : To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = $1.66054 X 10^{-24} g$

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X $1.66054 X 10^{-24} g/amu$ = $1.673 X 10^{-24} g$

Now, Volume = 1/6πd³ as diameter is given as $1.0 X 10^{-15} m$ converting it to cm will require to multiply with 100

∴ Volume = 1/6π $(1.0 X 10^{-15}mX 100 cm / 1 m)^{3}$

Hence, volume = $5.236 X 10^{-40} cm^{3}$

Therefore, Density = mass / volume

∴ Density = $1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}$

Therefore, Density will be $3.2 X 10^{15} g/cm^{3}$ .

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