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3 years ago

What volume of .100 M HCl is required to titrate .100 g of sodium bicarbonate to its equivalence point?

Chemistry

2 answers:

Art [367]3 years ago

8 0

Answer:

agree with what the person below me said!

Explanation:

Karo-lina-s [1.5K]3 years ago

6 0

The reaction is: HCl + Na HCO3 → NaCl + H2CO3, which is already balanced.

Then the molar ratios are 1 mol HCL : 1 mol NaHCO3

You know the mass of NaHCO3 is 0.100 g, then you can find the number of moles using the molar mass of NaHCO3

molar mass of NaHCO3 is 23 g/mol + 1 g/mol + 12g/mol + 3*16g/mol = 84 g/mol

And the number of moles is: 0,100 g / 84 g/mol = 0,0012 mol of NaHCO3.

Then, from the ratio 1:1, you know that, at the titration point, the number of moles of HCl is the same: 0,0012 mol of HCl.

Now from the fomula of molarity you have: M = #of moles / V in liters

=> V in liters = # of moles / M = 0.0012 mol / 0.100 M = 0.012 liters

Answer: 0.012 liters of HCL

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Rama09 [41]

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3 years ago

A prairie dog eats grass. A coyote eats the prairie dog. This is an example

Oduvanchick [21]

Answer:

This is an example of a food chain

Explanation:

Think of it as a chain reaction. The grass feeds and nourishes the prairie dog. Upon eating the prairie dog, the coyote gets the nutrients from both the grass the prairie dog ate and from the prairie dog itself.

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3 years ago

Balance the following equation in acidic conditions. Phases are optional. Cu NO3,

Gwar [14]

If your equation is Cu+NO^- 3-->Cu^2+NO, then the answer is
2 Cu + 1 NO3{-} → 1 Cu^{2+} + 3 NO

To check if it is balance, this is the solution:
2- Cu- 2
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5 0

3 years ago

If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera

Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0

3 years ago

Please answer with an actual answer and not just put a random word ^-^

WARRIOR [948]

Which eclipse was modeled when the large ball was between the small ball and the light?

The model is a "Lunar Eclipse" (If it was talking about the earth, then yes, it is a lunar eclipse).

Which eclipse was modeled when the small ball was between the large ball and the light?

The model is a "Solar Eclipse".

What does the large ball represent?

The earth.

What does the small ball represent?

The moon.

What does the light source represent?

The sun.

Hope this helps!~ <3

(I can't draw so sorry.)

6 0

3 years ago