Answer:
in 6.0 E3 g of blood at a concentration of 0.62 ppm there are 3.5g of lead content
Explanation:
- blood density: 1.06 g/L
- ppm ≡ mg Pb / L blood
⇒ V blood = 6.0E3gblood * L / 1.06gblood = 5660.37 L blood
<u> mass of lead:</u>
⇒ 5660.37 L blood * 0.62 mg Pb / L blood * 0.001 g / mg
⇒ mass = 3.509 g of lead
I'm assuming you mean how many grams of sodium does 8.99 grams of sodium bromide have, because Bromine would not have any sodium in it.
Then, just direct calculate 22.34
Explanation:
Relation between entropy change and specific heat is as follows.

The given data is as follows.
mass = 500 g,
= 24.4 J/mol K
= 500 K,
= 250 K
Mass number of copper = 63.54 g /mol
Number of moles = 
= 
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
= ![7.86 \times 24.4 \times [500 -T_{f}]](https://tex.z-dn.net/?f=7.86%20%5Ctimes%2024.4%20%5Ctimes%20%5B500%20-T_%7Bf%7D%5D)

= 750
So,
= 
- For the metal block A, change in entropy is as follows.

= ![24.4 log [\frac{375}{500}]](https://tex.z-dn.net/?f=24.4%20log%20%5B%5Cfrac%7B375%7D%7B500%7D%5D)
= -3.04 J/ K mol
- For the block B, change in entropy is as follows.

= ![24.4 log [\frac{375}{250}]](https://tex.z-dn.net/?f=24.4%20log%20%5B%5Cfrac%7B375%7D%7B250%7D%5D)
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.
Answer:
A. Stable
Explanation:
unless ur talking about the star in your PFP, then it's unstable xd
(a) sodium hydroxide + ammonium sulfate --> sodium sulfate + ammonium hydroxide
sodium sulfate precipitates in aqueous solutions
(b)<span> niobium(V) sulfate + barium nitrate → Barium sulfate + niobium nitride
both do not precipitate in aqueous solution
</span>(c) strontium bromide + silver nitrate ---> strontium nitrate + silver bromide
both do not precipitate in aqueous solution