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2 years ago

Bruh Im 4th on the leaderboard How is this possible

Chemistry

1 answer:

forsale [732]2 years ago

4 0

Answer:

I guess you just answered a lot of questions

Explanation:

Thanks for the points btw :)

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What volume of a 0.452 m naoh solution is needed to neutralize 85.0 ml of a 0.176 m solution of h2so4?

ArbitrLikvidat [17]

H2SO4 + 2 NaOH ----> Na2SO4 + 2 H2O.

0.085 L * 0.176 mol/L = 0.01496 mol H2SO4

is neutralised by 0.01496 mol * 2

= 0.02992 mol NaOH.

1000 mL of 0.492 M NaOH

contains 0.492 moles NaPH.

0.02992 / 0.452 * 1000 mL

= 66.19 = 66 mL

6 0

2 years ago

Read 2 more answers

The work function of an element is the energy required to remove an electron from the surface of the solid. The work function fo

Murrr4er [49]

Answer:

$\lambda=249.2\ nm$

Explanation:

Given that:

The work function of the rhodium = 480.5 kJ/mol

It means that

1 mole of electrons can be removed by applying of 480.5 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 480.5 kJ of energy.

1 electron can be removed by applying of $\frac {480.5}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $79.78\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $79.78\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{79.78\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:79.78}$

$\lambda=\frac{19.878}{10^6\times \:79.78}$

$\lambda=2.4916\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=249.2\ nm$

6 0

3 years ago

When 16.9 g KOH is dissolved in 90.8 g of water in a coffee-cup calorimeter, the temperature rises from 18.5 °C to 34.27 °C. Wha

juin [17]

Answer:

-431.5 J/g

Explanation:

Mass of solution = Mass of solute + mass of solvent

Solute is KOH while solvent is water.

Mass of KOH = 16.9 g

Mass of water = 90.8 g

Mass of solution = 16.9 + 90.8

= 107.7 g

Change in temperature (Δt) = 34.27 - 18.5

= 16.2 °C

Heat required to raise the temperature of water is released by dissolving KOH.

Therefore,

Heat released by KOH = m × s× Δt

= 107.7 × 4.18 × 16.2

= 7293 J

Heat released by per g KOH = 7293 J/16.9 g

= 431.5 J/g

As heat is released therefore, enthalpy change would be negative.

Enthalpy change of KOH = -431.5 J/g

4 0

2 years ago

What is the value of x A. 76 B. 56 C. 46 D. 66

vladimir1956 [14]

It’s d I’m pretty sure

7 0

2 years ago

A(n) 7.88-mol sample of carbon monoxide was stored in a 30.0-L container at 49.4°C. What is the

kirill115 [55]

The answer would be c I think k

7 0

2 years ago