All categories

2 years ago

If you sprayed an air freshener, where would the highest concentration of air freshener be found?

Chemistry

1 answer:

Hunter-Best [27]2 years ago

7 0

FLOOR. Most of the molecules that dont sick to something fall on the floor so most would be on the FLOOR.

You might be interested in

In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g

amm1812

The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O

number of moles = mass / molar mass
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams

3 0

3 years ago

Read 2 more answers

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

What is the volume of an oxygen tank if it contains 12 moles of oxygen at 273 K under 75 kPa?

Zina [86]

Given:
n = 12 moles of oxygen
T = 273 K, temperature
p = 75 kPa, pressure

Use the ideal gas law, given by
$pV=nRT \\ or \\ V= \frac{nRT}{p}$
where
V = volume
R = 8.3145 J/(mol-K), the gas constant

Therefore,
$V= \frac{(12\,mol)(8.3145\, \frac{J}{mol-K} )(273\,K)}{75 \times 10^{3} \, Pa}= 0.3632\,m^{3}$

Answer: 0.363 m³

5 0

3 years ago

What were the number of sunspots for the last 5 solar maximums?

Dennis_Churaev [7]

Solar Cycle 25 is now underway and expected to peak with 115 sunspots in July 2025.

8 0

2 years ago

What is the difference between a quantity and a unit?

Irina-Kira [14]

$Question$

What is the difference between a quantity and a unit?

Answer:

Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements. Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four fundamental units.

Hope this helps!!!

$xXxAnimexXx$

4 0

2 years ago