Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³
Therefore, we have;
The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Answer:
goes very fast and carries astronauts
Explanation:
Answer: combustion causes a chemical reaction between nitrogen and oxygen in the engine.
Explanation:
Nitrogen oxides are produced in combustion processes, partly from nitrogen compounds in the fuel, but mostly by direct combination of atmospheric oxygen and nitrogen in flames. Nitrogen oxides are produced naturally by lightning, and also, to a small extent, by microbial processes in soils.
Answer:
D. 6.00 L
Explanation:
What we have here is an example of Boyle's Law. The equation here is P₁ · V₁ = P₂ · V₂. We know all of the values except for V₂.
60(8) = 80V
<em>Multiply 60 by 8 to get 480.</em>
480 = 80V
<em>Divide both sides by 80.</em>
480/80 = V
6 = V
The final volume for the gas is 6.00 L.
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution
=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution
2) Molarity = M = number of moles of solute / liters of solution
=> calculate the number of moles of 8.9 grams of NaCl
3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol
4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol
5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M
Answer: 0.152 M
