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2 years ago

Help me guys please i need help

Chemistry

1 answer:

bearhunter [10]2 years ago

7 0

Answer:

It helps microorganisms grow

Explanation:

I think this is the correct answer because I tried searching it up and you sterilise the petri dish before adding the agar, I've used agar in petri dishes a year ago and im pretty sure they don't make cells glow or make bacteria appear larger.

if I had to have a second guess I'd go with the first one but considering the fact that bacteria is usually to small to see with the naked eye the last option seems correct

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What is the correct IUPAC name for NH4CL?

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Ammonium chloride is the correct name for NH4CL

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A piston in a heat engine does 500 joules of work, and 1,400 joules of heat are added to the system. Determine the change in int

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The change is that the piston gets hotter? (Theres more heat in it)

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3 years ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

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3 years ago

Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g

Lerok [7]

The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.

The reaction for the decomposition of water has been:

$\rm 2\;H_2O\;\rightarrow\;H_2\;+\;O_2$

From the balanced equation, 2 moles of water decomposes to form 1 moles of hydrogen and 1 mole of oxygen.

The mass of oxygen produced has been 50 g. The moles of oxygen has been given by:

$\rm Moles=\dfrac{mass}{molar\;mass}$

The moles of oxygen has been:

$\rm Moles_O_2=\dfrac{50}{32}\;mol\\Moles_O_2=1.5625\;mol$

The moles of oxygen produced has been 1.5625 mol.

The moles of hydrogen decomposed has been given from the balanced chemical equation as:

$\rm 1 \;mole\;O_2=2\;mole\;H_2O\\1.5625\;mol\;O_2=1.5625\;\times\;2\;mol\;H_2O\\1.5625\;mol\;O_2=3.125\;mol\;H_2O$

The moles of hydrogen decomposes has been 3.125 mol.

The mass of hydrogen decomposed has been given by:

$\rm Mass=moles\;times\;molar\;mass\\Mass_{H_2O}=3.125\;\times\;18.01\;g\\Mass_{H_2O}=56.28\;g$

The mass of water decomposed to produce 50 g oxygen has been 56.28 g. Thus, option D is correct.

For more information about moles produced, refer to the link:

brainly.com/question/10606802

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