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Nady [450]
3 years ago
11

Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen

ce point?
Chemistry
1 answer:
alexgriva [62]3 years ago
8 0

Answer:

V_{base}=24.04mL

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

2M_{base}V_{base}=M_{acid}V_{acid}

Therefore, for is to compute the volume of the used base, we proceed as shown below:

V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}

And we plug in to obtain:

V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL

Best regards!

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Ilia_Sergeevich [38]

27) Partial pressure of oxygen: 57.8 kPa

29) Final volume: 80 mL

30) Final volume: 8987 L

31) Due to property of water of being polar, ice floats on water

Explanation:

27)

In a mixture of gases, the total pressure of the mixture is the sum of the partial pressures:

p_T = p_1 + p_2 + ... + p_N

In this problem, the mixture contains 3 gases (helium, carbon dioxide and oxygen). We know that the total pressure is

p_T=201.4 kPa

We also know the partial pressures of helium and carbon dioxide:

P_{He}=125.4 kPa\\P_{CO_2}=18.2 kPa

The total pressure can be written as

p_T=p_{He}+p_{CO_2}+p_{O_2}

where p_{O_2} is the partial pressure of oxygen. Therefore, we find

p_{O_2}=p_T-p_{He}-p_{CO_2}=201.4-125.4-18.2=57.8 kPa

29)

Assuming that the pressure of the gas is constant, we can apply Charle's law, which states that:

"For an ideal gas at constant pressure, the volume of the gas is proportional to its absolute temperature"

Mathematically,

\frac{V}{T}=const.

where

V is the volume of the gas

T is the Kelvin temperature

We can re-write it as

\frac{V_1}{T_1}=\frac{V_2}{T_2}

Here we have:

V_1 = 42 mL (initial volume)

T_1=-89^{\circ}C+273=184 K is the initial temperature

T_2=77^{\circ}C+273=350 K is the final temperature

Solving for V2, we find the final volume:

V_2=\frac{V_1 T_2}{T_1}=\frac{(42)(350)}{184}=80 mL

30)

For this problem, we can use the equation of state for ideal gases, which can be written as

\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}

where in this problem:

p_1 = 102.3 kPa is the initial pressure

V_1=1975 L is the initial volume

T_1=25^{\circ}C+273=298 K is the initial temperature

p_2=21.5 kPa is the final pressure

T_2=12^{\circ}C+273=285 K is the final temperature

And solving for V2, we find the final volume of the balloon:

V_2=\frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(102.3)(1975)(285)}{(21.5)(298)}=8987 L

31)

A molecule of water consists of two atoms hydrogen bond with an atom of oxygen (H_2 O) in a covalent bond.

While the molecul of water is overall neutral, due to the higher electronegativity of the oxygen atom, electrons are slightly shifted towards the oxygen atom; as a result, there is a slightly positive charge on the hydrogen side, and a slightly negative charge on the oxygen side (so, the molecules is said to be polar).

As a consequence, molecules of water attract each other, forming the so-called "hydrogen bonds".

One direct consequence of the polarity of water is that ice floats on liquid water.

Normally, for every substance on Earth, the solid state is more dense than the liquid state. However, this is not true for water, because ice is less dense than liquid water.

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3-pentanone to form 3-pentanol

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In a 71.4 g sample of nahco 3 3 ​ , how many grams of sodium are present?
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There are 19.5 g Na in 71.4 g NaHCO₃

Calculate the <em>molecular mass of NaHCO₃</em>.

1 Na = 1 × 22.99 u = 22.99   u

1 H   = 1 × 1.008 u =    1.008 u

1 C   = 1 × 12.01  u =  12.01    u

3 O = 3 × 16.00 u =  <u>48.00  u </u>

               TOTAL =  84.008 u

So, there are 22.99 g of Na in 84.008 g NaHCO₃.

∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na

4 0
3 years ago
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