CHEMISTRY COLLEGE

For each set of dilutions (bleach and mouthwash), the first tube contained a 1:11 dilution (0.5 mL of agent was added to 5.0 mL of nutrient broth). What was the dilution of the MIC for the mouthwash and the bleach? Hint: You will have to calculate the dilution of each tube 2 – 4.

Answers

Answer 1

Answer:

The dilution of each tube are as follows;

Tube 2; 8.26 × 10-³
Tube 3; 7.5 × 10-⁴
Tube 4; 6.83 × 10-⁵

For each time a dilution is further diluted;

The dilution ratio is; 1 : 11; In essence, 0.5 mL of agent was added to 5.0 mL of nutrient broth.

For the first tube; dilution factor is; 1 : 11 = 9.1× 10-²

For the second tube; 0.5mL of the first tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) = 1/121 = 8.26 × 10-³.

For the third tube; 0.5mL of the second tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) ×(1/11) = 1/121 = 7.5 × 10-⁴.

For the fourth tube; 0.5mL of the third tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) × (1/11) × (1/11) = 1/14641 = 6.83 × 10-⁵.

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What nuclide undergoes fission to form molybdenum-103, atomic number 42, tin-131, atomic number 50, and two neutrons? Element symbol: _______ mass number: _______

Answers

Answer:

The correct answer is Pu, 234.

Explanation:

In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,

Mass of X = Mass of Molybdenum + Mass of Tin + Mass of neutrons

M = 1 * 103 + 1 * 131 + 2 * 0

M = 234

Now the atomic number (balanced) on both the sides,

Atomic number of X = Atomic number of Molybdenum + Atomic number of Tin + Atomic number of neutrons

A = 1*42 + 1*50 + 2*1

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The atomic number 94 is for the element Plutonium, whose symbol is Pu. Thus, the reactant is 234-Pu.

Most wine is prepared by the fermentation of the glucose in grape juice by yeast: C6H12O6(aq) --> 2C2H5OH(aq) + 2CO2(g) How many grams of glucose should there be in grape juice to produce 725 mLs of wine that is 11.0% ethyl alcohol, C2H5OH (d=0.789 g/cm3), by volume?

Answers

Answer:

123.41 g

Explanation:

Given that the ethyl alcohol produced is 11.0 % by volume.

It means that 1000 mL contains 110 mL of ethyl alcohol

Given that the volume is:- 725 mL

So,

Volume of ethyl alcohol = $(110)/(1000)* 725\ mL$ = 79.75 mL

Given that:- Density = 0.789 g/cm³ = 0.789 g/mL

So, Mass = Density*Volume = $0.789* 79.75\ g$ = 62.92 g

Calculation of the moles of ethyl alcohol as:-

Molar mass of ethyl alcohol = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = (Mass\ taken)/(Molar\ mass)$

Thus,

$Moles= (62.92\ g)/(46.07\ g/mol)$

$Moles=1.37\ mol$

According to the reaction:-

$C_6H_(12)O_6_((aq))\rightarrow 2C_2H_5OH_((aq)) +2CO_2_((g))$

2 moles of ethyl alcohol is produced when 1 mole of glucose reacts

Also,

1.37 moles of ethyl alcohol is produced when $(1)/(2)* 1.37$ mole of glucose reacts

Moles of glucose = 0.685 Moles

Molar mass of glucose = 180.156 g/mol

Mass = Moles*Molar mass = $0.685* 180.156\ g$ = 123.41 g

Please help me i don’t know the answer to this!!!!!!

Answers

Answer: Lithium and Ethanol

Explanation: As lithium and ethanol both have a density of under

1.000g/ml, they can be supported by the water. Blood and Magnesium are denser than the water and will therefore sink as the water can not support it them.

Which of the following required Bohr's model of the atom to need modification ? A. Energies of electrons are quantized. B. Quantized electron energies are responsible for emission spectra lines. C. An electron's energy increases the farther it moves from the nucleus. D. Electrons do not follow circular orbits around the nucleus..

Answers

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

Bohr's model of the atom is a combination of elements of quantum theory and classical physics in approaching the problem of the hydrogen atom. According to Neils Bohr, stationary states exist in which the energy of the electron is constant. These stationary states were referred to as circular orbits which encompasses the nucleus of the atom. Each orbit is characterized by a principal quantum number (n). Energy is absorbed or emitted when an electron transits between stationary states in the atom.

Sommerfeld improved on Bohr's proposal by postulating that instead of considering the electron in circular orbits, electrons actually orbited around the nucleus in elliptical orbits, this became a significant improvement on Bohr's model of the atom until the wave mechanical model of Erwin Schrödinger was proposed.

Answer:

Electrons do not follow circular orbits around the nucleus

Explanation:

How many grams of CO2 and H2O are produced from the combustion of 220. g of propane (C3H8)? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Answers

Answer:

(C3H8) produces 660 g of CO2 and 360 g of H2O

Explanation:

The balanced chemical equation for the combustion of propane (C3H8) is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

This equation tells us that for every molecule of propane (C3H8) that reacts with 5 molecules of oxygen (O2), 3 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O) are produced.

So, if we have 220. g of propane (C3H8), we can find the amount of CO2 and H2O produced by using the mole ratio from the balanced equation:

1 mole C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

We can find the number of moles of C3H8 by dividing the mass by the molar mass of C3H8 (44 g/mol):

220 g / 44 g/mol = 5 moles C3H8

So, the number of moles of CO2 and H2O produced can be found by multiplying the number of moles of C3H8 by the mole ratio:

3 moles CO2 = 3 moles CO2/1 mole C3H8 * 5 moles C3H8 = 15 moles CO2

4 moles H2O = 4 moles H2O/1 mole C3H8 * 5 moles C3H8 = 20 moles H2O

Finally, we can convert the number of moles of CO2 and H2O to grams by multiplying by their molar masses (44 g/mol for CO2 and 18 g/mol for H2O):

15 moles CO2 * 44 g/mol = 660 g CO2

20 moles H2O * 18 g/mol = 360 g H2O

So, the combustion of 220 g of propane (C3H8) produces 660 g of CO2 and 360 g of H2O.

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Answers

Answer:

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