Answer:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
Explanation:
The standard formation reaction is the synthesis of 1 mole of a substance from its elements in their most stables forms under standard conditions. The balanced chemical equation is:
Na(s) + C(s, graphite) + 1/2 H₂(g) + 3/2 O₂(g) → NaHCO₃(s)
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:
D is correct
Explanation:
because
we know that
density of lead is 11.36 g/cm3
and
density of tin is 7.31 g/cm3
so..
density of alloy by mixing 50/50
=(11.36+7.31)/2 g/cm3
=18.67/2 g/cm3
=9.33 g/cm3
Answer:
64567000000 nanolitres
Explanation:
Base 10 decimal system: 1 milli = 1000000 nano
We simply multiply 64,567 millilitres by 1000000 to get our number in nanolitres:
64567(1000000) = 64567000000 nanolitres
Answer:
V₂ = 104.76 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial temperature = 21°C (21 + 273.15 K = 294.15 K)
Final temperature = 35°C (35 + 273.15 K = 308.15 k)
Final volume = ?
Solution:
Charles Law:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ =100.0 mL × 308.15 K / 294.15 K
V₂ = 30815 mL.K /294.15 K
V₂ = 104.76 mL
