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3 years ago

Exactly 1 mole of alpo4 contains how many moles of al p and o

1 answer:

7 0

ALPO4 can be broken up into AL/P/O4. There is one AL mole, one P mole, and four O moles. Together, that is a total of six moles -- 1 + 1 + 4 = 6. You can find the moles for each section by breaking the atoms up by type. There are 4 O because the number 4 follows the O. There are none in front of AL or P, so they are singular.

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A 650.0 g hard-water sample contains 101 mg Ca.what is this concentration in parts per million

ki77a [65]

Answer:

= 155 ppm

Explanation:

PPM also refers to parts per million, it represents a low concentration of a solution. It represents 0.001 gram or a milligram in a 1000 mL, equivalent to 1 mg per liter

Given that;

101 mg of Ca in 650.0 g of water

1 ppm = 1 mg/L

650 g = 650 mL = 0.65 L

Therefore;

= 101 mg/ 0.65 L

= 155.38 mg/L

= 155 ppm

8 0

3 years ago

Balance the following Equation:

lisabon 2012 [21]

Answer:

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Option D is correct

Explanation:

Step 1: Data given

Mass of Zinc (Zn) = 50.0 grams

Mass of Hydrogen chloride (HCl) = 50.0 grams

atomic mass Zn = 65.38 g/mol

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol

Moles Zn = 0.764 moles

Moles HCl = 50.0 grams / 36.46 g/mol

Moles HCl = 1.37 moles

Step 4: Calculate limiting reactant

For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Zn is in excess. There will react 1.37/2 = 0.685 moles

There will remain 0.764 -0.685 = 0.079 moles

7 0

4 years ago

Which element has similar properties to Berylium? Explain

GaryK [48]

This has multiple answers can you be more specific

7 0

3 years ago

What is the mass of 4.5 moles of sodium fluoride?

Thepotemich [5.8K]

Mass of sodium is 23 and mass of fluirine is 19 there mass of NaF is 42
42 g = 1 mole therefor 4.5 moles will have
4.5 × 42 = 189 g

3 0

3 years ago

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

Answer: The average atomic mass of element Zinc is 65.40 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

For $_{30}^{67}\textrm{Zn}$ isotope:

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

For $_{30}^{68}\textrm{Zn}$ isotope:

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

For $_{30}^{70}\textrm{Zn}$ isotope:

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0

3 years ago