Answer:
2.49 × 10⁶ molecules
Explanation:
Given data
- Pressure (P): 9.25 × 10⁻¹⁴ atm
- Volume (V):

We can calculate the moles of gas using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 9.25 × 10⁻¹⁴ atm × 1.10 × 10⁻³ L / (0.0821 atm.L/mol.K) × 300.0 K
n = 4.13 × 10⁻¹⁸ mol
1 mole contains 6.02 × 10²³ molecules (Avogadro's number). The number of molecules in 4.13 × 10⁻¹⁸ moles is:
4.13 × 10⁻¹⁸ mol × (6.02 × 10²³ molecule/1 mol) = 2.49 × 10⁶ molecule
Answer: During the day, a typical greenhouse will trap Infrared photons from the sun, which allows the plants inside to stay warm at night. ... The much needed sunlight will still come through but the extra layer of protection will keep your plants safe at night
Explanation: Hope that helps :)
Answer:
490 in^3 = 8.03 L
Explanation:
Given:
The engine displacement = 490 in^3
= 490 in³
To determine the engine piston displacement in liters L;
(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)
First, we will convert in³ to cm³
Since 1 in = 2.54 cm
∴ 1 in³ = 16.387 cm³
If 1 in³ = 16.387 cm³
Then 490 in³ = (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³
∴ 490 in³ = 8029.63 cm³
Now will convert cm³ to dm³
(NOTE: 1 L = 1 dm³)
1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm
∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³
If 1 cm³ = 1 × 10⁻³ dm³
Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³
≅ 8.03 dm³
∴ 8029.63 cm³ = 8.03 dm³
Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³
Since 1L = 1 dm³
∴ 8.03 dm³ = 8.03 L
Hence, 490 in³ = 8.03 L
Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures?
</em>
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.