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jekas [21]
3 years ago
5

What is the correct name for this formula: AI2O3?

Chemistry
1 answer:
agasfer [191]3 years ago
5 0

Answer- A

Explanation- Because you have to look at the symbols and remember which symbol goes with each chemical, if you don't then you could get confused.

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A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coefficient is "1". KOH + Cu(NO3)2 → KNO
svet-max [94.6K]

2KOH + Cu(NO3)2 → 2KNO3 + Cu(OH)2

2K⁺ 1Cu²⁺ 2K⁺ 1Cu²⁺

2OH ⁻ 2NO3⁻ 2NO3⁻ 2OH⁻

5 0
3 years ago
A compound with an empirical formula of CH2O has a molar
NeTakaya

Answer:

c2h4o2

Explanation:

I am using my cell. This could take a bit

The Empical formula has a mass of

C 12

2H 2

O 16

total = 30

The molecular mass is given as 60 which is 2 times the empirical mass. Therefore multiply each element by 2 in the empirical formula.

You get C2H4O2

5 0
3 years ago
What is the hardest subject u ever taking, mine is chemistry
artcher [175]

Answer:

Mine is language arts brainliest?

Explanation:

7 0
3 years ago
Calculate the pressure of 2 mol of a gas at 300 K in 8 L container.
Sergio039 [100]

Answer:

4800

Explanation:

using my Cal ex to solve the question

calculation goes like this

2*300*8=4800

3 0
3 years ago
A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)
son4ous [18]

Answer:

\large \boxed{\text{4.5 atm}}

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?;          T₂ =  20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ =  (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}

6 0
3 years ago
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