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Svetllana [295]
3 years ago
6

I NEEEEEEEEEEEEEEEEEEEEEEEDDDDDDDDDDDDDDDDDD HEEEEEEEEEEELLLLLLLLLLLLPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Chemistry
2 answers:
olga2289 [7]3 years ago
8 0
Answer: 15g

Explanation:
vivado [14]3 years ago
4 0

Answer:

15g

Explanation:

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Determine the volume, in liters, of 3.2 mol of CO2 gas at STP.
Allushta [10]

Answer:

71.7 L

Explanation:

Using the ideal gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/Kmol)

T = temperature (K)

According to the information provided in this question;

P = 1 atm (STP)

V = ?

n = 3.2mol

T = 273K (STP)

Using PV = nRT

V = nRT/P

V = 3.2 × 0.0821 × 273/1

V = 71.7 L

6 0
3 years ago
Does anyone know the answer?????​
Juli2301 [7.4K]

Answer:

In chemistry, pH (/piːˈeɪtʃ/) (abbr. power of hydrogen or potential for hydrogen) is a scale used to specify how acidic or basic a water-based solution is. Acidic solutions have a lower pH, while basic solutions have a higher pH.

Explanation:

that should answer ur question

3 0
3 years ago
Students conduct an experiment:
sweet-ann [11.9K]

Answer:

The balloon becomes inflated

Explanation:

The equation of the reaction between baking soda (sodium bicarbonate) and vinegar(ethanoic acid) is shown below;

NaHCO3 + HC2CH3O2 ------> NaC2H3O2 + H2O + CO2

The gas (CO2) evolved in the process leads to the inflation of the balloon dropped on the bottle in which the reaction is taking pace.

This observation provides evidence that a gas was really evolved in the reaction.

7 0
3 years ago
A strong acid, HA, is added to water. Which statement about the solution is true?
Arturiano [62]
My answer to this question is C

3 0
3 years ago
A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r
svp [43]

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

8 0
3 years ago
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