Answer:
Destiny = 5 g/ml
Explanation:
The formula for density is d=
Knowing this, divide your mass by volume:
10 g / 2 ml = 5 g/ml
Final answer:
Destiny = 5 g/ml
Answer:
reactive nonmetals since they have a full valence shell (that's why they're stable).
The balanced chemical equation for the production of chromium metal from the reaction of chromium(ll) nitrate reacts with a strip of zinc is:
3 Zn + 2 Cr(NO₃)₃ → 2 Cr + 3 Zn(NO₃)₂
This is a redox reaction, which <u>is a chemical reaction in which one or more electrons are transferred between the reagents</u>, causing a change in their oxidation states. In the proposed reaction, Cr oxidation state goes from +3 to 0, becoming metallic chromium, while Zn goes from being Zn⁰ to Zn²⁺.
<u>The mass of chromium metal produced in the above reaction will be,</u>
425.0 mL x
x
x
x
= 5.52 g
So, the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess is 5.52 g of Cr.
Answer: m= 3.15x10-3 g NaHCO3
Explanation: To find the mass of NaHCO3 we will use the relationship between moles and molar mass. The molar mass of NaHCO3 is 84 g.
3.75x10-5 moles NaHCO3 x 84 g NaHCO3 / 1 mole NaHCO3
= 3.15x10-3 g NaHCO3
Answer:
(a)

(b)

Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:

Best regards.