All categories

2 years ago

How many grams of tungsten trioxide must you start with to prepare 1.80 g of tungsten? (For WO3, MW = 231.8 amu.)

2 answers:

5 0

In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.

A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass.

The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.

Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.

If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.

Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.

Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.

This page was loaded in 0.003 seconds.

Leona [35]2 years ago

4 0

6.43g. Tungsten is 183.84g/mol. 5.10g W / 183.84 g x mol = about .02774 mol. 1 mol of Tungsten trioxide makes i mol of tungsten, atomic weight of WO3 is 231.84 g/mol, multiply that by 0.02774 mol = 6.43g with significant digits.

You might be interested in

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.

6 0

2 years ago

Which of the following are examples of plasma?

artcher [175]

The answers are "tails of comets, the ionosphere, and a neon sign." or options A, D, and E. Tails of comments are mainly made of plasma. Ice cubes are examples of a solid. A gas fire key word gas is example of a gas. The ionosphere is also mainly made of plasma. A neon sign uses plasma in order to work. A flashlight uses a light bulb which isn't a example of plasma therefore thats a no.

Hope this helps!

4 0

2 years ago

Read 2 more answers

A chemist heats 100.0 g of FeSO4 x 7H2O in a crucible to drive off the water. If all the water is driven off, what is the mass o

Ierofanga [76]

FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)

M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol

m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)

m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)

m(FeSO₄)=151.9*100.0/278.0=54.6 g

m(FeSO₄)=54.6 g

4 0

2 years ago

Read 2 more answers

The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th

kenny6666 [7]

Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

For 725 mL of chlorine ≡ 0,725L

moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× $\frac{3molCl_{2}}{2 molCH_{4}}$ = 0,0198 moles

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = 0,0171 mol Cl₂

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% = 0,0102 moles of CH₄ And the moles that don't react are 0,00300 mol

Thus, moles of each compound are:

0,0102 moles of CH₄× $\frac{3Cl_{2}}{2 molCH_{4}}$ = 0,0153 mol + 0,0171 mol = 0,0324 mol Cl₂

0,0102 moles of CH₄× $\frac{2CH_{3}Cl}{2 molCH_{4}}$ = 0,0102 mol CH₄

0,0102 moles of CH₄× $\frac{2HCl}{2 molCH_{4}}$ = 0,0102 mol HCl

0,0102 moles of CH₄× $\frac{2Cl^{-}}{2 molCH_{4}}$ = 0,0102 mol Cl⁻

Total pressure using:

P = nRT/V

Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

-To obtain partial pressure you change the moles for each compound-

I hope it helps!

5 0

2 years ago

Determine the number of bonding electrons and the number of nonbonding electrons in the structure of xef2. enter the number of b

lakkis [162]

Answer : The number of bonding electrons and the number of non-bonding electrons are (4, 18).

Explanation :

The number of bonding electrons and non-bonding electrons in the structure of $XeF_2$ is determined by the Lewis-dot structure.

Lewis-dot structure : It tell us about the number of valence electrons of an atom within a molecule and it is also shows the bonding between the atoms of a molecule and the lone-pair of electrons.

In the given structure, 'Xe' is the central atom and 'F' is the terminal atom.

Xenon has 8 valence electrons and fluorine has 7 valence electrons.

Total number of valence electrons in $XeF_2$ = 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we conclude that

The number of electrons used in bonding = 4

The number of electrons used in non-bonding (lone-pairs) = 22 - 4 = 18

Therefore, the number of bonding electrons and the number of non-bonding electrons are (4, 18).

The Lewis-dot structure of $XeF_2$ is shown below.

4 0

2 years ago

Read 2 more answers