All categories

3 years ago

The atomic mass of titanium is 47.88 atomic mass units. This atomic mass represents the

1 answer:

3 0

Answer (3) is the most correct, although (2) is not to be ignored. (3) states the most abundant isotope Ti's average mass, which is certainly true. (2) is the total mass of all protons, neutrons, and electrons in an atom of Ti, which is true but has to be more specific in order to pinpoint exactly the 47.88 amu. (4) is incorrect because it is not of all the naturally occurring isotopes of Ti. (1) is incorrect because they forgot electrons.

You might be interested in

how many moles of sodium hydroxide would have to be added to 250 ml of a 0.403 m acetic acid solution, in order to prepare a buf

jonny [76]

The volume of the buffer solution having a ph value is calculated by henderson's hasselbalch equation.

Buffer solution is water based solution which consists of a mixture containing a weak acid and a conjugate base of the weak acid. or a weak base and conjugate acid of a weak base.it is a mixture of weak acid and a base. The pH of the buffer solution is determined by the expression of the henderson hasselbalch equation.

pH=pKa + log [salt]/[acid]

Where, pKa =dissociation constant , A- = concentration of the conjugate base, [HA]= concentration of the acid. Here, a buffer solution contains 0.403m acetic acid and 250 ml is added in order to prepare a buffer with a ph of 4.750. Putting all the values in the henderson hasselbalch equation we find the pH of the buffer solution.

To learn more about hendersons hasselbalch equation please visit:

brainly.com/question/13423434

#SPJ4

6 0

1 year ago

How does a buffer resist change in ph upon addition of a strong acid?

lora16 [44]

Any buffer exists in this equilibrium

HA <=> $H^+ + A^-$
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base ( $A^-$ )

When a strong acid is added, it reacts with the large reservoir of the conjugate base ( $A^-$ ) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.

8 0

3 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

2 years ago

I NEED ANSWER PLEEEAASSEE!!!

emmasim [6.3K]

The answer is B. This is because Sodium has 1 valence electron and Fluorine has 7 valence electrons. All elements want 8 valence electrons so they may be stale, like the noble gases are. Hope this helps.

3 0

3 years ago

Read 2 more answers

What is the volume occupied by 4.20 miles of oxygen gas (O2) at STP

Drupady [299]

Answer: 94.13 L

Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.

To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.

8 0

3 years ago