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3 years ago

4. What is the bridge that allows you to convert from the given substance to the wanted substance

1 answer:

6 0

I’m not sure if this is what you are referring to but there is something called a moles bridge which allows you to convert from one substance to another

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Calculate ∆U (the change in internal energy) for a system on which 850.00

Illusion [34]

Answer:

The heat gain by the system,

−

250

The work done on the system ,

−

500

The First Law of Thermodynamics state that

−

750

Explanation:

6 0

2 years ago

PLZ HELP ME !!! PLZ. What is N for silicon tetrachloride, SiCl4?

den301095 [7]

Answer:To find S, the number of shared

electrons, recall that

S = N-A

Explanation:UwU

4 0

3 years ago

Isotopes percent abundance of antimony

Nitella [24]

Antimony has two naturally occurring isotopes. Their abundance is given in the pic attached below

4 0

3 years ago

A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

$pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )$

$pH = 9.25$

8 0

3 years ago

Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come

Elden [556K]

Answer:

Explanation:

For solving this we need a heat balance

$Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}$

By changing the corresponding relations, we have

$m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\$

By cancelling similar factor, we obtain

$\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\$

Which means that the change of temperature in A is twice the change of B

3 0

3 years ago