Answer:
4.57 × 10⁶ nanometers
Explanation:
In this problem we are asked to convert between two units i.e. a meter into nanometer. In sciences, the different units are used for a same quantity (achieved by multiplying conversion factors) to get rid of very small values and get a readable and intelligible values.
For Example:
In given statement the value of small distance is 4.5 × 10⁻³ meter the real number form of this number is 0.00457. Hence, this number can be converted to a very large number by multiplying it with 10⁹ or 1000000000. Hence,
4.5 × 10⁻³ × 1.0 × 10⁹ = 4.57 × 10⁶
Or,
4.5 × 10⁻³ meters = 4.57 × 10⁶ nanometers
Although the quantity is the same for same units but the number has changed.
Answer:
d is the right one. aaßriadjcjfjfjfdjfjf
Answer:
Explanation:
For a general equilibrium
aA +bB ⇔ cC + dD ,
the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.
Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).
So with this in mind, lets answer this question.
1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.
2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.
3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.
4. True: From our previous reasongs this is the true one.
5. False: If K is small, the equilibrium lies on the reactants side.
Answer:
3 mol AlCl₃.
Explanation:
Hello!
In this case, according to the specified reactants and products, it is possible to set up the following balanced chemical reaction:
Whereas we evidence the 1:3 mole ratio between aluminum nitrate and sodium chloride; thus, since different moles were reacting, we need to identify the limiting reactant by computing the moles of AlCl₃ produced by each reactant as follows:
Thus, we infer that NaCl is the limiting reactant as it produces the fewest moles of AlCl₃; consequently the produced amount of this product is 3 mol.
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