Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of
= 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 2 mole of
react with 1 mole of 
So, 5 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is 
To solve this kinematics formula use the following equation:
Vf = Vi + at
Vf = 0 + (9.81 m/s^2)(3 seconds)
Vf = 29.43 m/s and or about 29.4 m/s of reported to 3 significant figures.
Molar mass H₂O = 18.0 g/mol
number of moles :
1.0 / 18.0 => 0.055 moles
1 mole -------------- 6.02 x 10²³ molecules
0.055 moles -------- ? molecules
molecules = 0.055 x ( 6.02 x 10²³) / 1
molecules = 3.311x10²² / 1
= 3.311 x 10²² molecules
hope this helps!
Answer:
3.676 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
- Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
Answer:
look at the graph
Explanation:
We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.
Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.
So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)