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3 years ago

What element is used in bright flashing advertising signs

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

4 0

The element used in advertising signs is neon.

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11) If you have a density of 100 kg/L,, tell me the following:

jenyasd209 [6]

The unit of mass is 'Kilogram' which is written as 'kg' and volume, v = 10 L.

<h3>Equation :</h3>

To calculate the volume

Use formula,

density = mass / volume

density = 100 kg/L

mass = 1000 kg

volume = mass / density

v = 1000/100

v = 10 L

<h3>What is density mass?</h3>

A substance, material, or object's mass density is a measure of how much mass (or how many particles) it has in relation to the volume it occupies.

To know more about volume :

brainly.com/question/1578538

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I understand the question you are looking for :

If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units? Secondly, what is the volume?

8 0

1 year ago

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

6 0

3 years ago

Help, I need to convert 3.16 μm to inches!

irina1246 [14]

0.00011811023622 <span>i</span><span>nches</span>

7 0

3 years ago

Please answer as (a;3) < example>

never [62]

A;4, b;6, c;7, d;5, e;8, f;3

7 0

3 years ago

How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1

valentinak56 [21]

<span>4 Al + 3 O2 → 2 Al2O3

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>

7 0

3 years ago

Read 2 more answers