Answer: there is more atoms of each element on the right.
Explanation:
On the left of the arrow you have:
Si-1
F-4
H-4
O-2
On the right of the arrow you have:
Si-3
F-6
H-8
O-4
A balanced equation always has the same amount of each element on the left of the arrow (the reactants) as the right (the products). This isn’t balanced because you have a lot more of each element on the right.
The fact that they lay like 300 eggs. And maybe that they are fast.
Answer:
1/3p0
Explanation:
The combined gas law:
P1V1/T1 = P2V2/T2, where P, V and T are Pressure, Volume, and Temperature. Temperature must always be in Kelvin. The subscriopts 1 and 2 are for initial (1) and final (2) conditions.
In this case, temperature is constant (adiabatically). V1 = 2.0L and V2 = 6.0L. I'll assume P1 = p0.
Rearrange the combined gas law to solve for final pressure, P2:
P1V1/T1 = P2V2/T2
P2 = P1*(V1/V2)*(T2/T1) [Note how I've arranged the volume and temoperature terms - as ratios. This helps us understand what the impact of raising or lowering one on the variables will do to the system].
No enter the data:
P2 = P1*(V1/V2)*(T2/T1): [Since T2 = T1, the (T2/T1) terms cancels to 1.]
P2 = p0*(2.0L/6.0L)*(1)
P2 = (1/3)p0
The final pressure is 1/3 the initial pressure.
Given :
Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .
To Find :
How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .
Solution :
By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.
So , volume of solution does not matter .
Moles of oxygen , 
.
Now , molecule of CO contains 1 mole of C .
So , moles of C is also 0.167 mole .
Mass of carbon , 
.
Therefore , mass of carbon is 2 grams .
Hence , this is the required solution .
