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3 years ago

I NEED HELP PLEASE! :)

Chemistry

1 answer:

Dmitrij [34]3 years ago

5 0

Answer:

Explanation:

SF4 forms a trigonal bipyramidal shape and its molecular shape is that of a "see-saw".

Since sulfur is in 3rd period, it violates the octet rule and has more than 8 electrons accompanying its valence shell.

Here's a picture of lewis structure and electron geometry.

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If a volume of gas 177.6mL was collected at a temperature of 25.8C and the pressure is 799.7 torr, what is the original concentr

Step2247 [10]

Answer:

The original concentration of the acid was 0.605 M

Explanation:

Step 1: Data given

Volume of gas = 177.6 mL = 0.1176 L

Temperature = 25.8 °C = 298.95 K

Pressure = 799.7 torr = 799.7/ 760 = 1.0522368 atm

Volume of acid needed to react = 12.6 mL = 0.0126 L

Step 2: Calculate moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the number of moles = TO BE DETERMINED

⇒with p = the pressure of the gas = 799.7 torr = 1.0522368 atm

⇒with V = the volume of the gas = 177.6 mL = 0.1776 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 25.8 °C = 298.95 K

n = 0.007618 moles

Step 3: Calculate original concentration

We need 0.007618 moles of acid to react with the same amount of moles gas

Concentration acid = moles / volume

Concentration acid = 0.007618 moles / 0.0126 L

Concentration acid = 0.605 M

The original concentration of the acid was 0.605 M

5 0

3 years ago

A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

Ostrovityanka [42]

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

4 0

3 years ago

Which of the following best describes technology

kotegsom [21]

were are the choices

5 0

3 years ago

Read 2 more answers

In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc

natita [175]

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water

x= 0.7535 g water

3 0

3 years ago

The process by which waste products are eliminated from the cell by diffusion through the cell membrane is called _____.

madam [21]

Answer:

Exocytosis

Explanation:

8 0

2 years ago