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3 years ago

Pls help!!!!!

Chemistry

2 answers:

umka2103 [35]3 years ago

7 0

The exact positions of the electrons in an atom cannot be determined because electrons are always .

erica [24]3 years ago

3 0

Answer:

moving i think! Hope this helps

Explanation:

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What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a

Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K [At STP]

Now first find the number of moles

F₂ + CaBr₂ → CaF₂ + Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So, 199.89 g CaBr₂ reacts with = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

$n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}$

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

#SPJ4

4 0

1 year ago

Historically the troubadour was a ____.

eimsori [14]

Composer and performer of Old Occitan lyric poetryduring the High Middle Ages (1100–1350). Since the word troubadour is etymologically masculine, a female troubadour is usually called a trobairitz.

6 0

3 years ago

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a

DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0

3 years ago

Compound element ratio of 1:2:6<br> Ba(NO3)2+Na2SO4

Karo-lina-s [1.5K]

1:2:6? Ba(NO3)2 +Na2SO4 -----> BaSO4 + 2NaNO3

Ba(NO3)2

BaSO4

NaNO 3

Na2SO4

5 0

3 years ago

Read 2 more answers

What is the summary at the end of an experiment that explains the results

jasenka [17]

Answer:

The conclusion

Explanation:

6 0

3 years ago