The solution of this problem with explaining the solution?

Choose the correct name for each acid. Click here

Zigmanuir [339]

Answer:

1. Nitrous Acid

2. Hydroiodic Acid

3. Phosphoric Acid

Explanation:

Just did the lesson

7 0

3 years ago

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Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0

3 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

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

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

The activation energy for the decomposition = 33813.28 J/mol

8 0

3 years ago

I WILL GIVE BRAINLIEST PLEASE DO IT RIGHT!!! 1) Using the data below, create two simple line graphs on a separate sheet of paper

VLD [36.1K]

Answer:Score” scatter plot shows an example of a positive relationship—as one variable increases, so does the other. The points in this type of scatter plot tend to go “uphill” from left to right

Explanation:Score” scatter plot shows an example of a positive relationship—as one variable increases, so does the other. The points in this type of scatter plot tend to go “uphill” from left to right

3 0

3 years ago

4 Compared to the energy and charge of the electrons in the first shell of a Be atom, the electrons in the second shell of this

Korvikt [17]

Moving the electron away from the nucleus requires energy, so the electrons in the outer shell will have more energy than ones in the inner shell. Electrons always have a charge of -1, so the charge in the inner and outer shell will be the same. Therefore the answer is 3

5 0

4 years ago

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