Find concentration of each ions: 600.0 ml of solution containing 1.50 grams of Cr(NO3)2

Match each term to its description. (3 points)

madreJ [45]

Answer: Limiting reactant = 3

Theoretical Yield= 1

Excess reactant=2

Explanation: The theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction. It can be calculated from: the balanced chemical equation. the mass and relative formula mass of the limiting reactant , and. the relative formula mass of the product.

An excess reactant is a reactant present in an amount in excess of that required to combine with all of the limiting reactant. It follows that an excess reactant is one remaining in the reaction mixture once all the limiting reactant is consumed.

The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated

3 0

3 years ago

Read 2 more answers

Why is carbon dioxide called a gas and not vapour?

Afina-wow [57]

Explanation:

CO2 is called as gas because it exist in single thermodynamics state i.e CO exist in gases state only at room temperature.

5 0

2 years ago

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

Answer:

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.

7 0

2 years ago

Select all statements that correctly describe the reactions of benzene. A. Benzene typically undergoes reactions in which the ar

Katarina [22]

Answer:Benzene typically undergoes reactions in which the aromatic ring is preserved.B. Benzene typically reacts with electrophiles where an aromatic proton is substituted by the electrophile

Explanation:

The reactions of benzene are such that the aromatic ring is not destroyed. Addition reactions destroy the aromatic ring hence they aren't typical reactions of benzene. Benzene rings are attacked by electrophiles in which reaction a proton is substituted by the electrophile. Alkenes only undergo addition reaction and not electrophilic substitution reaction.

8 0

3 years ago

Hydrogen (H), oxygen (O), and Nitrogen (N) are all examples of which of the following?

Llana [10]

Answer:

All are correct

Explanation:

This might be a little deceptive. The question shows H, O and N which would designate ELEMENTS.

However, all of these can also be considered compounds and molecules as well. BUT, they are H₂, O₂ and N₂ when they are molecules and compounds.

So it depends on the whether the question is being literal or not.

8 0

2 years ago