How many grams of fluorine are needed to generate 7.65 moles Carbon Tetrafluoride

When 5.467 grams of compound z are burned in excess oxygen, 15.02 grams of co2 and 2.458 grams of h2o are produced. determine it

Vera_Pavlovna [14]

Given the mass of CO2 =15.02 g

Moles of $CO_{2}$ = $15.02 gCO_{2} *\frac{1 molCO_{2} }{44.01 gCO_{2} } =0.341 mol CO_{2}$

Mass of H2O = 2.458 g

Moles of $H_{2}O$ = $2.458 g H_{2}O*\frac{1molH_{2}O }{18.02g H_{2}O } =0.136molH_{2}O$

Moles of C = $0.341 mol CO_{2}*\frac{1 molC}{1 molCO_{2} } =0.341molC$

Moles of H = $0.136 mol H_{2}O * \frac{2 mol H}{1 mol H_{2}O } =0.272 mol H$

Mass of C in the sample = $0.341 mol C*\frac{12.01g C}{1 mol C} = 4.095 g C$

Mass of H = $0.272 mol H *\frac{1.01 g H}{1 molH}$ =0.275 gH

Mass of O in the sample = 5.467 g - (4.095 g +0.275 g) = 1.097 g O

Moles of O = $1.097 g O * \frac{1 molO}{16 g O} =0.0686 mol O$

Simplest mole ratios of the elements in the compound:

$Cx_{\frac{0.341mol}{0.0686mol}} H_{\frac{0.272mol}{0.0686mol} }O_{\frac{0.0686mol}{0.0686mol} }$

Therefore the empirical formula of the compound is $C_{5}H_{4}O$

7 0

3 years ago

Lewis dot diagram for hydrogen

Oksanka [162]

· H

Lewis dot diagrams only represent the valence electrons the element contains, and Hydrogen only has one valence electron.

7 0

3 years ago

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HELP HELP ASAP<br> I WILL MARK BRAINLIEST

likoan [24]

Answer:

i cant read it

Explanation:

6 0

3 years ago

Chlorine has two natural isotopes, chlorine-35 and chlorine-37.

Alex Ar [27]

Answer:

The relative atomic mass of chlorine depends on the ratio between the abundance of these two naturally-occurring isotopes.

Explanation:

The relative atomic mass of an element is a weighted average of the atomic mass of its naturally-occurring isotopes. The relative abundance of each isotope gives its weight in this weighted average.

For these two naturally-occurring isotopes of chlorine:

The relative atomic mass of $^{35}{\rm Cl}$ is approximately $34.969$ daltons. The relative abundance of this isotope in nature is approximately $0.758$ .
The relative atomic mass $^{37}{\rm Cl}$ is approximately $36.966$ daltons. The relative abundance of this isotope in nature is approximately $0.242$ .

$\begin{array}{|c|c|c|}\cline{1-3} \text{Isotope} & \text{Atomic Mass} & \text{Relative Abundance}\\ \cline{1-3} ^{35}{\rm Cl} & \approx 34.968\; \rm Da} & \approx 0.758 \\ \cline{1-3} ^{37}{\rm Cl} & \approx 36.966\; \rm Da & \approx 0.242 \\ \cline{1-3}\end{array}$ .

$\begin{aligned}&\text{relative atomic mass of Cl} \\ &= \text{atomic mass of $^{35}{\rm Cl}$} \times \text{relative abundance of $^{35}{\rm Cl}$} \\&\quad + \text{atomic mass of $^{37}{\rm Cl}$} \times \text{relative abundance of $^{37}{\rm Cl}$} \\ &\approx 34.968\; \rm Da \times 0.758 + 36.966\; \rm Da \times 0.242 \\ &\approx 35.45\; \rm Da \end{aligned}$ .

The relative abundance of $^{35}{\rm Cl}$ is much higher than that of $^{37}{\rm Cl}$ . Consequently, the relative atomic mass of the element $\rm Cl$ is closer to the atomic mass of $^{35}{\rm Cl}\!$ than that of $^{37}{\rm Cl}\!$ .

7 0

3 years ago

Doing an Endothermic reaction, what happens to energy in relation to the surroundings?

natali 33 [55]

Answer:

Doing an Endothermic reaction, energy is absorbed from surroundings

Explanation:

Endothermic reactions:

The type of reactions in which energy is absorbed are called endothermic reactions.

In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.

For example:

C + H₂O → CO + H₂

ΔH = +131 kj/mol

it can be written as,

C + H₂O + 131 kj/mol → CO + H₂

we can see that 131 kj/mol energy is taken by the reactants. So energy is absorbed from surrounding.

Exothermic reaction:

The type of reactions in which energy is released are called exothermic reactions.

In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.

For example:

Chemical equation:

C + O₂ → CO₂

ΔH = -393 Kj/mol

it can be written as,

C + O₂ → CO₂ + 393 Kj/mol

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3 years ago

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