Answer:
Concentric circles of contour lines indicate a hilltop; concentric circles with hachure marks indicate a close depression.
Explanation:
Hope this helped Mark BRAINLIEST!!!
Answer : The rate for a reaction will be 
Explanation :
The balanced equations will be:
In this reaction, 
and 
are the reactants.
The rate law expression for the reaction is:
![\text{Rate}=k[A]^2[B]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D%5E1)
or,
![\text{Rate}=k[A]^2[B]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D)
Now, calculating the value of 'k' by using any expression.
Now we have to calculate the initial rate for a reaction that starts with 1.48 M of reagent A and 1.32 M of reagents B.
![\text{Rate}=k[A]^2[B]^0[C]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5E2%5BB%5D%5E0%5BC%5D%5E1)
Therefore, the rate for a reaction will be
Answer:
2.174 gm
Explanation:
PV = nRT n = number of moles
R = gas constant = .082057 L-atm/(K-mol)
T must be in units of K
.870 (3.95) = n (.082057)(35+273.15)
solve for n = .1359 moles
Methane mole weight (CH4) = 16 gm / mole
.1359 moles * 16 gm/mole = 2.174 gm
BOBERT don’t u dare write a whole paragraph cause no one has time to read all that rn xoxo
Answer:
7.5 g of AlCl3
Explanation:
The given equation is;
NaOH + AlCl3 --> Al(OH)3 + NaCI.
By inspection, it is not balanced because OH and Clare not equal on both sides of the equation.
Thus, let's make them equal by balancing the equation.
Cl has 3 on the left, so we will make it to have 3 on the right. Same thing with OH on the right and we will make it to have 3 on the left. Thus:
3NaOH + AlCl3 --> Al(OH)3 + 3NaCI
We can see that;
NaOH has 3 moles
While AlCl3 has 1 mole
Thus, to find how many grams of AlCl3 will be required to completely react with 2.25g of NaOH ;
2.25g of NaOH × (3 moles NaOH/39.997 g/mol of NaOH) × (1 mole of AlCl3/3 moles of NaOH) × (133.34 g/mol of AlCl3/1 mol AlCl3) = 7.5 g of AlCl3
