Someone please help me dont scroll i need help

If the initial concentration of a is 0.00540 m, what will be the concentration after 795 s?

zavuch27 [327]

Feb 19, 2014 - The units of k tell you that this is a second order reaction. So, to solve this, you need to use the integrated rate law for a 2nd order reaction: 1/[A] = kt + 1/[A]o 1/[A] = 0.540/Ms (835 s) + 1/0.00640 1/[A] = 607 [A] = 1.65X10^-3 M.

7 0

3 years ago

(((NEED ANSWER QUICK!!!))) Which is the stronger conjugate base, CN- or OCN-? Explain

Shkiper50 [21]

Answer:

The stronger conjugate base will be the weaker acid; i.e., the acid with the smaller Ka-value.

Explanation:

Given conjugate base CN⁻ => weak acid => HCN => Ka =4.9 x 10⁻¹⁰

Given conjugate base OCN⁻ => weak acid=> HOCN => Ka = 3.5 x 10⁻⁴

Ka(HCN) << Ka(HOCN) => CN⁻ is a much stronger conjugate base than OCN⁻

6 0

3 years ago

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C

laila [671]

Answer:

First choice: 99 J/K

Explanation:

1) First law of thermodynamic (energy balance)

Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

2) Energy change of each substance:

General formula:

Heat released or absorbed = mass × Specific heat × change in temperature

density of water: you may take 0.997 g/ ml as an average density for the water.

mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

Specif heat of water: 1 cal / g°C

Heat released by the hot water:

Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

Heat absorbed by the cold water:

Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

Heat absorbed by the calorimeter

Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

4) Balance

Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

Convert to cal / K to Joule / K

1 cal = 4.18 Joule

23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

4 0

4 years ago

What is the concentration of H+ ions at a pH = 11?

natta225 [31]

Answer:

$\huge 1 × {10}^{-11} \: \: M$

Explanation:

The pH of a solution can be found by using the formula

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

Since we are finding the H+ ions we find the antilog of the pH

So we have

$11 = - log({H}^{+}) \\ {H}^{+} = {10}^{ - 11}$

We have the final answer as

$1 × {10}^{-11} \: \: M$

Hope this helps you

6 0

3 years ago

Consider the following reaction: 2Mg(s)+O2(g)-->2MgO(s) delta H=-1204kJ

Pachacha [2.7K]

Answer:

a. The reaction is exothermic.

b. -87,9 kJ

c. 9,60g of Mg(s)

d. 602kJ are absorbed

Explanation:

Based on the reaction:

2Mg(s) + O₂(g) → 2MgO(s) ΔH = -1204kJ

a. The reaction is exothermic. Because ΔH<0. That means the reaction produces heat when occurs

b. 3,55g of Mg(s) are:

3,55g Mg × ( 1mol / 24,305g) = 0,146 moles of Mg(s)

As 2 moles of Mg(s) produce -1204 kJ of heat:

0,146 moles of Mg(s) × ( -1204kJ / 2mol Mg) = -87,9 kJ

c. If -238 kJ of heat were transferred. The moles of Mg(s) that react must be:

-238kJ × ( 2mol Mg / -1204kJ) = 0,395 moles of Mg(s). In grams:

0,395 moles × ( 24,305g / 1mol Mg) = 9,60g of Mg(s)

d. The reverse reaction is:

2MgO(s) → 2Mg(s) + O₂(g) ΔH = +1204kJ

40,5g of MgO(s) are:

40,5g MgO × ( 1mol MgO / 40,3044g) = 1,00 moles of MgO(s)

As 2 moles of MgO absorbe 1204kJ of energy:

1,00 moles of MgO(s) × ( +1204 kJ / 2mol MgO) = 602kJ are absorbed

I hope it helps!

7 0

3 years ago