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2 years ago

2 QUESTIONS: ANSWER THIS with a proper answer plz. PLZ DO NOT ANSWER JUST FOR MY BRAINLY POINTS!!!!

Chemistry

1 answer:

leonid [27]2 years ago

3 0

Answer:

Si hay algo que sucedió entre ustedes raro o un comportamiento que tuviste con esa persona o que te vio hacer asi como le pudieron decir algo de ti que no le gustara a tu amigo hace que se comporte raro contigo

Explanation:

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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.

6 0

2 years ago

Giving brainiest to whoever answers correctly.

Digiron [165]

Answer:250

Explanation:

5 0

2 years ago

Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

2 years ago

2. ATO kg bouting ball would require what force to accelerate down an alleyway at a rate of ons?

Mars2501 [29]

Answer:

2. 30N

3. 5,000N

4. 15 kg

5. 2,800 kg

Explanation:

2. A 10 kg bowling ball would require what force to accelerate down an alleyway at a rate of 3m/s² ?

Notice that I completed the question with the garbled and missing values:

Data:

F = ?

m = 10 kg

a = 3m/s²

Physical principles:

Newton's second law: $F=m\times a$

Solution:

Substitute and compute

$F=10kg\times 3m/s^2=30N$

3. Salty has a car that accelerates at 5 m/s². If the car has a mass of 1000 kg, how much force does the car produce?

Notice that I arranged the typos.

Data:

F = ?

m = ?

a = ?

Physical principles:

Newton's second law: $F=m\times a$

Solution:

Substitute and compute

$F=1,000kg\times5m/s^2=5,000N$

4. What is the mass of a falling rock if it produces a force of 147 N?

Data:

F = 147N

m = ?

a = falling rock

Physical principles:

neglecting air resistance ⇒ a = g: gravitational acceleration: 9.8m/s²

Newton's second law: $F=m\times a$

Solution:

Clear m from Newton's second law

$m=\dfrac{F}{a}$

Substitute with F = 147 N and a = g = 9.8m/s², and compute

$m=\dfrac{147N}{9.8m/s^2}=15Kg$

5. What is the mass of a truck if it produces a force of 14,000 N while accelerating at a rate of 5 m/s²?

Data:

F= 14,000N

m = ?

a = 5m/s²

Physical principles:

Second Newton's law: $F=m\times a$

Solution:

Clear m from Newton's second law

$m=\dfrac{F}{a}$

Substitute with F = 14,000 N and a = 5m/s², and compute

$m=\dfrac{14,000N}{5m/s^2}=2,800kg$

6 0

2 years ago

How many grams are there in 100L of H2 gas?

Ahat [919]

Answer:

1 mole =22.4 L of gas at STP. So now 1 mole of H2 gas weighs equal to 2 grams.

Explanation:

4 0

2 years ago

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