Please helppp :( According the VSEPR Theory, what shape is the "SiH3F" molecule?

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

4 years ago

Fill in the blank. The amount which an object accelerates

KATRIN_1 [288]

I think the answer is force

6 0

3 years ago

Write the electronic configuration of the following element or ion and circle on their valance electrons

agasfer [191]

1s2,2s2,2p6

Explanation:

electron arrangements

6 0

3 years ago

A student has 70.5 mL of a 0.463 M aqueous solution of sodium bromide. The density of the solution is 1.22 g/mL. Find the follow

AleksandrR [38]

Answer:

a.) 86.01 g.

b.) 3.36 g.

c.) 0.394 m ≅ 0.40 m.

d.) 4.77%.

e.) 3.9%.

Explanation:

a.) mass of the solution:

The density of the solution is the mass per unit volume.

∵ Density of solution = (mass of solution)/(volume of the solution).

∴ Mass of the solution = (density of solution)*(volume of the solution) = (1.22 g/mL)*(70.5 mL) = 86.01 g.

b.) grams of sodium bromide :

Molarity (M) is defined as the no. of moles of solute dissolved per 1.0 L of the solution.

∵ M = (no. of moles of NaBr)/(Volume of the solution (L))

∴ no. of moles of NaBr = M*(Volume of the solution (L)) = (0.463 M )*(0.0705 L) = 0.0326 mol.

∵ no. of moles of NaBr = (mass of NaBr)/(molar mass of NaBr)

∴ mass of NaBr = (no. of moles of NaBr)*(molar mass of NaBr) = (0.0326 mol)*(102.894 g/mol) = 3.36 g.

c.) molality of the solution:

Molality (m) is defined as the no. of moles of solute dissolved per 1.0 kg of the solvent.

∵ m = (no. of moles of NaBr)/(mass of the soluvent (kg))

no. of moles of NaBr = 0.0326 mol,

mass of solvent = mass of the solution - mass of NaBr = 86.01 g - 3.36 g = 82.65 g = 0.08265 kg.

∴ m = (no. of moles of NaBr)/(mass of the soluvent (kg)) = (0.0326 mol)/(0.08265 kg) = 0.394 m ≅ 0.40 m.

d.) % (m/v) of the solution:

∵ (m/v)% = [(mass of solute) /(volume of the solution)]* 100

∴ (m/v)% = [(3.36 g)/(70.5 mL)]* 100 = 4.77%.

e.) % (m/m) of the solution:

∵ (m/m)% = [(mass of solute) /(mass of the solution)]* 100

∴ (m/m)% = [(3.36 g)/(86.01 g)] * 100 = 3.9 %.

4 0

3 years ago

What is the frequency of UV light that has an energy of 5.2 x 10^-8 J?

adelina 88 [10]

The answer for this is because 98,8

3 0

3 years ago