To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
Answer:
if an atom gains an electron, the ion has negetive charge
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.
