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nikitadnepr [17]
3 years ago
5

How many molecules are there in 4.27 mol of tungsten(VI) oxide, WO3

Chemistry
1 answer:
Simora [160]3 years ago
8 0

Answer:

<h3>Molar mass of WO3 = 231.8382 g/mol </h3><h3> </h3><h3>Convert grams Tungsten(VI) Oxide to moles  or  moles Tungsten(VI) Oxide to grams </h3><h3> </h3><h3>Molecular weight calculation: </h3><h3>183.84 + 15.9994*3</h3>

Explanation:

<h3>#hopeithelps</h3><h3 /><h2>brainliest me later.</h2>
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3. Using the solubility of ionic compounds table and/or the solubility rules,
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Answer:

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Module 2 dba for chemistry questions? flvs
NeX [460]
Can you translate to english?

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4 years ago
predict where an unknown element that has these properties could fit in the periodic table 62 protons and electrons
Vesnalui [34]
Position of element in periodic table is depend on the electronic configuration of element.

Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.

Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.

Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table. 
5 0
3 years ago
NEED HELP FAST!!
koban [17]

Answer:

I'm pretty sure that's right.

Explanation:

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8 0
3 years ago
The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
s344n2d4d5 [400]

Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

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4 years ago
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