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2 years ago

What is the chemical formula for the compound tetraphosphorus decasulfide?

Chemistry

1 answer:

marusya05 [52]2 years ago

5 0

P S
4 10

you just use the roots given in the molecular formula

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When heating liquid materials in laboratory glassware, a student should always

Lelechka [254]

Answer:

be careful to not hurt yourself

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3 years ago

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How many pounds in a 3.00 l bottle of drinking water?

slava [35]

Answer:

6.61 Pounds

Solution:

Step 1: Calculate Mass of Water as;

Density = Mass ÷ Volume

Solving for Mass,

Mass = Density × Volume ------ (1)

As,

Density of Water = 1 g.cm⁻³

And,

3 L of Water = 3000 cm³

Putting values in equation 1,

Mass = 1 g.cm⁻³ × 3000 cm³

Mass = 3000 g

Step 2: Convert Grams into Pounds;

As,

1 Gram = 0.002204 Pounds

So,

3000 Grams = X Pounds

Solving for X,

X = (3000 Grams × 0.002204 Pounds) ÷ 1 Gram

X = 6.61 Pounds

6 0

3 years ago

What is the volume of 500g of CO2?

Airida [17]

Weighs 0.001836 gram per cubic centimeter or 1.836 kilogram per cubic meter

Try to see if this helps

3 0

3 years ago

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Is water wet? or is it just an experience?

polet [3.4K]

WATER is wet to make it a more marketable commodity

7 0

3 years ago

Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10

PtichkaEL [24]

Answer:

$K.E.=1.97\times 10^{-21}\ J$

Explanation:

Given that:-

Pressure = $6.9\times 10^5\ Pa$

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

Given the value of pressure = 43,836 Pa

So,

$6.9\times 10^5\ Pa$ = $\frac{6.9\times 10^5}{101325}$ atm

Pressure = 6.80977 atm

Volume = $2.3\times 10^{-3}\ m^3$ = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

$K.E.=\frac{3}{2}\times K\times T$

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

T is the temperature

So, $K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J$

$K.E.=1.97\times 10^{-21}\ J$

3 0

2 years ago