All categories

2 years ago

The photograph shows one way individuals work together in groups. What

Chemistry

2 answers:

slega [8]2 years ago

6 0

I think you forgot to upload the picture so I don’t think we can help you... sorry

siniylev [52]2 years ago

5 0

Answer:

the answers is administration for apxs

Explanation:

You might be interested in

When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.

sertanlavr [38]

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $15.8J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m$ = mass of water = 100.0 g

$\Delta T$ = change in temperature = $T_2-T_1=(32.0-23.9)=8.1^oC$

Now put all the given values in the above formula, we get:

$q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]$

$q=3513.8J=3.5138kJ$ (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole$

Now,

$\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol$

Therefore, the enthalpy change for the solution is 42.8 kJ/mol

4 0

3 years ago

Match each gas law with the correct statement.

Serhud [2]

The correct match are as follows:

Boyles Law is about a gas at constant temperature. Therefore, the correct match would be the volume of a gas varies inversely with pressure if the temperature is kept constant.

Charles Law would be that the volume of a gas varies directly with temperature if the pressure is kept constant.

Gay-lussac's law would be that the pressure of a gas varies directly with temperature if the volume is kept constant.

4 0

3 years ago

Read 2 more answers

Cyclobutane, C4H8, consisting of molecules in which four carbon atoms form a ring, decomposes, when heated, to give ethylene. Th

frosja888 [35]

Answer:

The concentration of cyclobutane after 875 seconds is approximately 0.000961 M

Explanation:

The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M

The final concentration of cyclobutane, [ $A_t$ ] = 0.00119 M

The time for the reaction, t = 455 seconds

Therefore, the Rate Law for the first order reaction is presented as follows;

$\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }$

Therefore, we get;

$k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }$

Which gives;

$k = \dfrac{\text{ ln} \dfrac {0.00119}{0.00150}} {-455} \approx 5.088 \times 10^{-4}$

k ≈ 5.088 × 10⁻⁴ s⁻¹

The concentration after 875 seconds is given as follows;

[ $A_t$ ] = [A₀]· $e^{-k \cdot t}$

Therefore;

[ $A_t$ ] = 0.00150 × $e^{5.088 \times 10^{-4} \times 875}$ = 0.000961

The concentration of cyclobutane after 875 seconds, [ $A_t$ ] ≈ 0.000961 M

6 0

3 years ago

A buffer contains significant amounts of ammonia and ammonium chloride. part a write an equation showing how this buffer neutral

siniylev [52]

Hello!

When HI is added, the buffer reacts in the following way:

1) Neutralizing of the Acid:

HI + NH₄OH → NH₄I + H₂O

2) Dissociation of the salt of a weak acid:

NH₄I → NH₄⁺(aq) + I⁻ (aq)

3) Dissociation of a weak acid to form H₃O⁺ (very little):

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

This series of reactions show how the adding of a strong Acid can be neutralized by the buffer, releasing instead very little amounts of Hydronium ions.

Have a nice day!

8 0

3 years ago

941 milliliters is the same as: cm3 and L

Nutka1998 [239]

For this case, we must take into account the following conversion factors:

$1 mL = 1 cm ^ 3$