All categories

3 years ago

Help me ples HELP HELP HELP HELP HELP

Chemistry

1 answer:

Darya [45]3 years ago

3 0

Answer:

Explanation:autotroph and photosynthesis

sorry if I am wrong I am dumb. :( :(

I found it on google.

God bless, stay safe, and good luck! :) :)

You might be interested in

How many grams of barium sulfate form when 35.0 ml of 0.160 m barium chloride reacts with 58.0 ml of 0.065 m sodium sulfate? do

vazorg [7]

Barium sulfate weighs 0.88 grams per gram.

Mass:

A physical body's overall composition is measured by its mass.

Inertia, or the body's resistance to acceleration in the presence of a net force, is also measured by it.

The mass of a thing also affects how strongly it attracts other bodies through gravity.

In the SI, the kilogram serves as the basic mass unit (kg).

This information

Moles of BaCl2 equal 5.6 x 103 (35 x 0.160 / 1000)

sodium sulfate moles equal 58 x 0.065, or 3.77 x 103,

Na2SO4 with BaCl2 ——————————-> 2NaCl plus BaSO4

Na2SO4 is the limiting agent here.

Therefore, the product relies on limiting reagent.

1 mol of Na2SO4 produces 1 mol of BaSO4.

0.00377 moles of BaSO4 is one mole.

moles = mass / molar mass of BaSO4.

0.00377 = mass / 233.4

BaSO4 mass is 0.88 g.

So, 0.88 grams of barium sulfate make up one gram.

To know more about the mass visit:

brainly.com/question/28204586

#SPJ4

3 0

1 year ago

When might neutralization reactions be used in a laboratory setting?

Ipatiy [6.2K]

Neutralization reactions can be used in a laboratory setting in order t<span>o dispose of chemicals. When spills happens, for instance an acid is on the floor, you can use a base to neutralize the spill. Hope this answers the question. Have a nice day.</span>

6 0

2 years ago

A 0. 462 g sample of a monoprotic acid is dissolved in water and titrated with 0. 180 m koh. what is the molar mass of the acid

Len [333]

The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.

<h3>For the calculation of molarity of solution</h3>

Molarity = (moles of solute/volume of solution) × 1000

Given,

Molarity of KOH solution = 0.180 M

Volume of solution = 28.5 mL

0.180 = (moles of KOH/ 28.5) × 1000

Moles = (0.18× 28.5)/1000

= 0.00513 mol

<h3>Chemical equation for the reaction</h3>

HA + KOH ------- KA + H2O

1 moles of KOH reacts with 1 moles of HA.

So, 0.00513 moles of KOH react with 0.00513 moles of HA.

<h3>To calculate the molar mass for given number of moles</h3>

Number of moles= given mass/ Molar mass

Given,

Mass of HA = 0.462 g

Moles of HA = 0.00512 mol

0.00512 = 0.462/ Molar mass

Molar mass = 90.23 g/ mol.

Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.

learn more about molar mass or moles:

brainly.com/question/26416088

#SPJ4

3 0

1 year ago

A sample of 8.5 g NH 3 on oxidation produces 4.5 g of NO. Calculate the percent yield.

yanalaym [24]

Answer:

nvnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnv

Explanation:

3 0

2 years ago

Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E

Hatshy [7]

Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

4 0

3 years ago