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Alex787 [66]
2 years ago
14

SOICHIOMETRY HOW MANY MOL OF NITROGEN DIOXIDE ARE REQUIRED TO PRODUCE 2.75 MOL NITRIC ACID? 3NO2+ H20→ 2HNO3+ NO

Chemistry
1 answer:
pishuonlain [190]2 years ago
4 0

Answer: C

Explanation: Hope this help :D

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A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is
Reika [66]

<u>Answer:</u> The pH of the buffer is 4.61

<u>Explanation:</u>

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.70

[\text{conjuagate base}]} = moles of conjugate base = 3.25 moles

[\text{acid}]  = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61

Hence, the pH of the buffer is 4.61

8 0
3 years ago
Cc
zzz [600]

Answer:

You should be posting this question under the biology tab, but here's the answer nonetheless.

a. 25% (if the woman is Ce)

c. 0% (if the woman isn't Ce)

5 0
3 years ago
If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f
I am Lyosha [343]

Answer:

0.032 L or 32 mL

Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

V2 = 2 L

Solve for V1 --> V1 = M2V2/M1

V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L

3 0
3 years ago
Ca3(PO4)2 + 2 H2SO4 = 2 CaSO 4 +Ca(H2PO4)2 is this balanced or unbalanced
ale4655 [162]

yes, it is balanced.

3 0
2 years ago
An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its
Aleks04 [339]

Answer:

T_1=-91.18\°C

Explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

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

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

T_1= \frac{T_2V_1}{V_2}\\\\T_1= \frac{(18.00+273.15)K(1.75L)}{(2.80L)}\\\\T_1=182K-273.15\\\\T_1=-91.18\°C

Best regards!

6 0
3 years ago
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