What is the correct formula for calcium fluoride

Hi :) can someone answer this thank u

scoray [572]

Add all the carbon atoms (3) then add hydrogen atoms (8)
C3H8

3 0

3 years ago

One of the hopes for solving the world's energy problem is

lions [1.4K]

Answer:

2.803013439419911 × 10⁻¹² J

Explanation:

Mass defect = mass of reactant - mass of product

(2.0140 + 3.01605) - (4.002603 + 1.008665)

5.03005 - 5.011268 = 0.018782 amu

mass in Kg = mass (amu) × 1.66053892173 × 10⁻²⁷ kg

mass in kg = 0.018782 × 1.66053892173 × 10⁻²⁷ = 3.1188242027932 × 10⁻²⁹kg

E = Δm c² where c is the speed of light = 2.9979 × 10⁸m/s

E = 3.1188242027932 × 10⁻²⁹kg × (2.9979 × 10⁸m/s)² = 2.803013439419911 × 10⁻¹² J

6 0

3 years ago

The reaction of C4H8 and Br2 to yield C4H2Br2 represents

natima [27]

Answer:

A Synthesis Reaction

Explanation:

A + B --> AB

8 0

3 years ago

Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

3 years ago

Keep your work area neal and

Sindrei [870]

Clean? I’m pretty sure not sure what it means but.

8 0

2 years ago