The standard on which the atomic mass unit is based is the mass of a (1 point)proton. neutron. chlorine-35 atom. carbon-12 atom.

Answers

Answer 1
Answer: D. carbon-12 atom I hope this helps :)
Answer 2
Answer:

Carbon-12 Atom is correct :)


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Calculate the mass of water produced when 2.06 g of butane reacts with excess oxygen.

Answers

2C4H10 + 13O2 = 8CO2 + 10H2O

1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10

2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O

3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O

A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.What is the percent yield of oxygen in this chemical reaction?

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The balanced chemical reaction will be:

2KClO3 = 2KCl + 3O2

 

We are given the amount of potassium chlorate being burned. This will be our starting point.

400.0 g KClO3 1 mol  KClO3/ 122.55  g KClO3) (3 mol O2/2 mol KClO3) ( 32.00 g O2/1mol O2) = 156.67 g O2

 

Percent yield = actual yield / theoretical yield x 100

 

Percent yield =115.0 g / 156.67  g x 100

Percent yield = 73.40 %

Calculate the pH of an acetic acid solution, originally 0.25 M, in water. The pKa for acetic acid = 4.76. Use the x-is-small approximation. Enter your answer with two digits after the decimal point.

Answers

Answer:

Initial concentration of acetic acid (CH3COOH_initial): 0.25 M

pKa for acetic acid: 4.76

Assume x is the concentration of H+ ions formed through dissociation.

CH3COOH ⇌ x (due to dissociation)

Apply the x-is-small approximation: We assume x is much smaller than the initial concentration of acetic acid (0.25 M). Therefore, we can neglect x in comparison to 0.25 M.

Calculate pH using the pKa equation:

Rounded to two decimal places, the pH of the acetic acid solution is approximately 2.68.

Explanation:

Which phrase describes the molecular structure and properties of two solid forms of carbon, diamond and graphite?(1) the same molecular structures and the same properties(2) the same molecular structures and different properties
(3) different molecular structures and the same properties
(4) different molecular structures and different properties

Answers

Answer is: (4) different molecular structures and different properties.

Different forms of the same element that have different properties because of different atom arrangements are called allotropes.  

Carbon has many allotropes, but two most important are graphite and diamomd.  

Graphite has sp2 and diamond has sp3 hybridization of carbon atoms, because of that graphite conduct electricity and diamond not.  

In diamond carbon atoms are arranged in the face centered cubic crystal structure called a diamond lattice.  

Diamond has very strong covalent bonds between carbon atoms and because of that it has the highest hardness and thermal conductivity of any bulk substance.  

The answer is (4) different molecular structures and different properties. The graphite has the structure of layer and diamond has the structure of regular tetrahedron.

Determine the new pressure of o2(g) in the cylinder in atmospheres

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It is an ideal gas therefore we can use the ideal gas equation to solve the problem. The ideal gas equation is expressed as PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature. V, n, and T  should be known to solve this problem.

Calculate the heat energy for this scenario: You are repeating the food experiment we did in class with a different food snack. At the beginning of this experiment, you started with 25 grams of water at 22 deg * C At the end of the experiment, the final temperature of the water is 45°C. The specific heat of water is 4.18J / (deg) * CSub and solve

Answers

Answer:

Mass of water (m) = 25 grams = 0.025 kg (since 1 g = 0.001 kg)

Specific heat of water (c) = 4.18 J/(g°C) = 4.18 J/(kg°C)

Initial temperature (T_ {initial}) = 22°C

Final temperature (T_(final) )= 45°C

Change in temperature (ΔT):

ΔT=T_(final)-T_ {initial}=45°−22°=23°

Now, calculate the heat energy (Q)

Q=mass×specific heat×ΔT

Q=0.025kg×4.18J/(kg°C)×23°C

Q≈2.44kJ

So, the heat energy for this scenario is approximately 2.44 kilojoules (kJ).