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Deffense [45]
3 years ago
13

Do you have 11 protons? Cause you're sodium fine.

Chemistry
1 answer:
mars1129 [50]3 years ago
3 0
Hey nice jeans there, fermi-cutie B))
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SO3 is an empirical formula for which of the following?
tino4ka555 [31]

Answer:

what are the options

Explanation:

there are no options

4 0
3 years ago
O2 would not be considered as a compound.<br><br> True<br> False
MissTica
<span>True, a compound consists of two different molecules/atoms</span>
4 0
4 years ago
The plastics industry realized more than 20 years ago that recycling plastics was a necessity for our planet. Can you recognize
Leni [432]

Answer:

C.)Recycled bottles and containers are used to produce lumber products that can be used as an outdoor building material.

Explanation:

4 0
3 years ago
Consider a piece of gold jewelry that weighs 9.85 g and has a volume of 0.675 cm3 . The jewelry contains only gold and silver, w
AlekseyPX

Answer:

61.5096 %

Explanation:

Let the amount of gold in the jewelry = x g

Let the amount of silver in the jewelry = y g

Total mass of jewelry = 9.85 g

So,

x + y = 9.85 g    .................................1

Density = Mass / Volume

Given that:

Density of gold = 19.3 g/cm³

Thus, volume is:

Volume = Mass / Density = x / 19..3 cm³

Given that:

Density of silver = 10.5 g/cm³

Thus, volume is:

Volume = Mass / Density = y / 10.5 cm³

The total volume of the jewelry = 0.675 cm³

So,

\frac {x}{19.3}+\frac {y}{10.5}=0.675

Or,

10.5 x + 19.3 y = 136.78875  ................2

Solving 1 and 2, we get that:

<u>x = 6.0587 g</u>

<u>y = 3.7913 g</u>

So, mass % of gold:

<u>Mass % = (6.0587 / 9.85)×100 = 61.5096 %</u>

7 0
3 years ago
Determine the volume (in liters) of a 87.8-9 sample of OF2 gas at 307 torr and 264°C.
expeople1 [14]

Answer:

V = 177.4 L.

Explanation:

Hello there!

In this case, since this gas can be assumed as ideal due to the given data, we can use the following equation:

PV=nRT\\\\PV=\frac{m}{MM}RT

Thus, by solving for volume we obtain:

V=\frac{mRT}{MM*P}

So we can plug in the temperature in Kelvins (537 K), the pressure in atmospheres (0.404 atm) and the molar mass (54 g/mol) to obtain:

V=\frac{87.8g*0.08206\frac{atm*L}{mol*K}*537K}{54g/mol*0.404atm}\\\\V=177.4L

Best regards!

6 0
3 years ago
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