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3 years ago

How many moles are in 15 grams of calcium

2 answers:

5 0

Atomic mass Ca = 40.0 u.m.a

1 mole Ca ------------- 40.0
? moles Ca ---------- 15 g

15 x 1 / 40 = 0.375 moles of Ca

chubhunter [2.5K]3 years ago

3 0

Approximately 0.374 moles

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In the reaction between p-aminophenol and acetic anhydride to form acetaminophen, 0.450 mL of water was added. What was the purp

BabaBlast [244]

Answer:

In the reaction between p-aminophenol and acetic anhydride to form acetaminophen, 4.5 ml of water were added. This was done to recover the product out the container. Water was used as a means of carrying out the product out the container. However, it should be noted that we would want a small amount to be added in order to avoid the product dissolving onto it.

Explanation:

4 0

2 years ago

What is the molar mass of a substance?

Sveta_85 [38]

Answer:

The mass in grams of one mole of a substance.

3 0

3 years ago

g Suppose a sample of an ideal gas in a container is subjected to a temperature change. A decrease in temperature will the kinet

Umnica [9.8K]

Answer:

323.15 °C

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.

P ∝ T

Also,

Using Charle's law

$\frac {P_1}{T_1}=\frac {P_2}{T_2}$

Given ,

P₂ = 2P₁

T₁ = 25 °C

T₂ = ?

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

Using above equation as:

$\frac{P_1}{298.15}=\frac{2P_1}{T_2}$

$T_2=2\times 298.15\ K$

New temperature = 596.3 K

Also,

T(K) - 273.15 = T( °C)

<u>So, Temperature = 596.3 - 273.15 °C = 323.15 °C</u>

5 0

3 years ago

An analytical chemist is titrating of a solution of ethylamine with a solution of . The of ethylamine is . Calculate the pH of t

Elenna [48]

Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

$C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+$

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

$n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol$

Thus, the concentration of ethylamine in solution is:

$[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M$

Now, we can also infer that some salt is formed, and has the following concentration:

$[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M$

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

$pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0$

Finally, the pH turns out to be:

$pH=14-pOH=14-3\\\\pH=11$

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

4 0

3 years ago

What do colligative properties depend on? A) the identity of the solute b) the number of particles present in the solution c) th

Nastasia [14]

B.) Colligative properties depend on <span>the number of particles present in the solution.

Hope this helps!</span>

4 0

2 years ago

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