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3 years ago

Match each property to the appropriate subatomic particle.

Chemistry

1 answer:

sladkih [1.3K]3 years ago

4 0

Proton = +1
neutron = 0
electron = -1

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How much work is required to lift a 0.500 kg block 18.5 m?

kipiarov [429]

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

3 0

3 years ago

What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?

VLD [36.1K]

K5O2

convert grams to moles, divide both by the smallest mole mass, multiply that until hole.

30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2

5 0

3 years ago

Pls help

Andrej [43]

Answer:

• None of these

Explanation:

your hand looks too old.

#Rastaman

5 0

2 years ago

NO + H2-> N2 + H2O<br> Balance equation?

Oksanka [162]

Answer:

2NO + 4H-> N2 + 2 H2O

Explanation:

Both sides must be equal. :)

5 0

3 years ago

What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor

ddd [48]

Answer:

$1.75272\ \text{m}^3$

Explanation:

The breakdown reaction of ozone is as follows

$CF_3Cl + UV \rightarrow CF_3 + Cl$

$Cl + O_3 \rightarrow ClO + O_2$

$O_3 + UV \rightarrow O_2 + O$

$ClO + O \rightarrow Cl + O_2$

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of $CF_3Cl$ = 15.5 g

M = Molar mass of $CF_3Cl$ = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = $62.363\ \text{L mmHg/K mol}$

Number of moles is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}$

$20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}$

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3$

For 20 cycles of the reaction the volume of the ozone is $1.75272\ \text{m}^3$ .

7 0

3 years ago