Answer:
20.0 grams
Explanation:
If the density of gold is 20.0 g/mL, then we can multiply it by 1 mLto find the weight of 1 mL of gold.
20.0
*1mL=20.0 grams
Air is matter bc it is a gas
Answer:
81.59%
Explanation:
First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:
- 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃
Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:
- 6.32 mol NH₃ *
= 6.32 mol NO
Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:
- 6.32 mol NO * 30 g/mol = 189.60 g NO
Finally we <u>calculate the percent yield</u>:
- 154.70 g / 189.60 g * 100% = 81.59%
Answer:
The answer is B- LiNO3(aq)+H2O(I)
Explanation:
I just did the assignment also UNUS ANNUS
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
