The boiling of water results in a physical change in matter form

What is an organism that survives on a hist organism and cuases harm to the host

umka21 [38]

An organisam is part of your body plant life

5 0

3 years ago

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

6 0

3 years ago

If the rate of decomposition of ammonia, NH3, at 1150 K is 2.10 x 10-6 mol/L/s, what is the

Alina [70]

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

3 0

3 years ago

By titration, it is found that 31.7 mL of 0.145 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentrati

arlik [135]

Answer:

0.184 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, the following data were obtained:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, the data obtained from the question. This includes:

Volume of the base, NaOH (Vb) = 31.7 mL Molarity of the base, NaOH (Mb) = 0.145 M

Volume of the acid, HCl (Va) = 25.0 mL

Molarity of the acid, HCl (Ma) =?

Finally, we shall determine the molarity of the acid (HCl) as shown below:

MaVa /MbVb = nA/nB

Ma × 25 / 0.145 × 31.7 = 1

Cross multiply

Ma × 25 = 0.145 × 31.7

Ma × 25 = 4.5965

Divide both side by 25

Ma = 4.5965 / 25

Ma = 0.184 M

Therefore, the molarity of the acid (HCl) is 0.184 M

6 0

3 years ago

Consider the following reaction and situations 1 through 10. In the spaces provided, clearly indicate the best response to each

olchik [2.2K]

Answer:

1. C. no change

2. A. increase

3. E. shift to the right

4. A. increase

5. E. shift to the right

6. A. increase

7. F. cannot be determined

8. B increase

9. D. shift to the left

10 F. cannot be determined

Explanation:

<em>According to Le Chaterlier principle, when a reaction is in equilibrium and one of the constraints that affect reactions is applied, the equilibrium will shift so as annul the effects of the constraints.</em>

From the equation: C(s) + H2O(g) ⇌ CO(g) + H2(g),

H is greater than 0, meaning that the system is endothermic, that is energy is absorbed.

1. If the pressure of the system is increased, there would be no change to the system because there are equal number of moles of products and reactants.

2. If H2 concentration is decreased, the equilibrium will shift to the right and more products will be formed. Hence, the concentration of CO will increase.

3. If H2 concentration is decreased, the equilibrium will shift to the right to annul the effects of the decrease in the concentration of a product.

4. If the concentration of H2 is increased, the equilibrium will shift to the left to annul the effects of increased concentration of a product. Hence, more H2O would be formed.

5. If H2 (a product) is removed, and C (a reactant) is added, more of the products will be formed in order to annul the effects of the actions. Hence, equilibrium will shift to the right.

6. If the amount of C (a reactant) is increased, the equilibrium will shift to the right. Hence, more H2 will be formed.

7. The reaction is endothermic, hence an increase in temperature will ordinarily shift the equilibrium to the right. However, the addition of H2 (a product) is supposed to shift the equilibrium to the left. Hence, the effects of simultaneous addition of the two actions become indeterminate.

8. Since the reaction is endothermic, increase in the temperature of the system will shift the equilibrium to the right. Hence, more CO will be formed.

9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

5 0

2 years ago