Ask question

Ask question

All categories

2 years ago

13

Determine ΔH for Reaction 4 from these data: Reaction 4: N2H4 (l) + 2H2O2 (l) → N2 (g) + 4H2O (l) ΔH 4 = ? Reaction 1: N2H4 (l)

+ O2 (g) → N2 (g) + 2H2O (l) ΔH 1 = -622.3 kJ Reaction 2: H2 (g) + 1/2O2 (g) → H2O (l) ΔH 2 = -285.8 kJ Reaction 3: H2 (g) + O2 (g) → H2O2 (l) ΔH 3 = -187.8kJ

1 answer:

kaheart [24]2 years ago

5 0

Answer:

2 Answers

Anonymous

7 years ago

Best Answer

Q = m*c*ΔT

Applying the first law of thermodynamics or the law of conservation of energy i.e

Energy lost by the metal == Energy gained by the water so now:

m1 * c1 *( Tinitial - Tfinal ) == m2 * c2 *( Tfinal- Tinitial )

(2.4)*(c1)*(98.88-20.60) == (28)*(4.2)*(20.60 - 19.73)

c1 = the specific heat of the unknown metal

c2 =the specific heat of water 4.2J/g K

Solve for c1

c1* 187.872 == 102.312

c1 == 0.5446 J/g ▫C

Note this:

1. Changing the temperature scale from Celsius to Kelvin is quite irrelevant here, as we are working with the difference only.

2. The interchanging of ( Tinitial - Tfinal) to *( Tfinal- Tinitial ), because of the substance is losing energy, yet while the other is gaining.

Explanation:

hope i helped : )

You might be interested in

The student is now told that the four solids, in no particular order, are calcium bromide (CaBr2), sugar (C6H12O6), benzoic acid

lara [203]

MgBr2, 3 ions per mole=best conductor

KBr, 2 ions per mole, 2nd best conductor

Benzoic acid, weak acid, slightly ionized, weak conductor

Sugar, molecular, non ionized, non conductor

8 0

2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago

In the chemical equation 3c2h4 how many atoms of carbon are represented

Scorpion4ik [409]

An example.
water is H2O

2 hydrogen, 1 oxygen

so the number to the right means how much of what is on the left.

so it looks like 2, because C2, but look at the 3 at the beginning. that means
3 (c2h4)

so 6 carbons, 12 hydrogen

the ratio of c2 to h4 doesn't change it's always 1:2.

but the 3 at the front is a different number relating to how much you have

5 0

2 years ago

What is the pH of a solution of 0.400 M CH₃NH₂ containing 0.250 M CH₃NH₃I? (Kb of CH₃NH₂ is 4.4 × 10⁻⁴)

Karolina [17]

Answer:

$\boxed{\text{10.84}}$

Explanation:

A solution of a weak base and its conjugate acid is a buffer.

The equation for the equilibrium is

$\rm CH$_3$NH$_2$ + H$_2$O $\, \rightleftharpoons \,$ CH$_3$NH$_2$+ H$_{3}$O$^{+}$\\\text{For ease of typing, let's rewrite this equation as}\\\rm B + H$_2$O $\longrightarrow \,$ BH$^{+}$ + OH$^{-}$; $K_{\text{b}}$ = 4.4 \times 10^{-4}$$

The Henderson-Hasselbalch equation for a basic buffer is

$\text{pOH} = \text{p}K_{\text{b}} + \log\dfrac{[\text{BH}^{+}]}{\text{[B]}}$

Data:

[B] = 0.400 mol·L⁻¹

[BH⁺] = 0.250 mol·L⁻¹

Kb = 4.4 × 10⁻⁴

Calculations:

(a) Calculate pKb

pKb = -log(4.4× 10⁻⁴) = 3.36

(b) Calculate the pH

$\text{pOH} = 3.36 + \log \dfrac{0.250}{0.400} = 3.36 + \log 0.625 = 3.36 - 0.204 = 3.16\\\\\text{pH} =14.00 -3.16 = \mathbf{10.84}\\\\\text{The pH of the solution is }\boxed{\textbf{10.84}}$

4 0

2 years ago

What volume would 0.435 moles of hydrogen gas, h2, occupy at stp?

faust18 [17]

The conversion factor for volume at STP is $\frac {1mol}{22.4L}$ or $\frac {22.4L}{1mol}$ . Since we want volume, we would use $\frac {22.4L}{1mol}$ . We conclude with the following calculations:

$0.435mol H_{2} * \frac {22.4LH_{2}}{1molH_{2}} = 9.744L$

The answer is 9.744L H2

3 0

2 years ago

Read 2 more answers