Answer: W + BgCz2 arrow WCz + Bg
2 W + BgCz2 arrow 2 WCz + Bg
Explanation:
Cz has 2 so you balcne the other side of WCz.
Since you Balcanes the Cz you changed the W and you Balcanes the other W on the left side.
The question is incomplete, here is the complete question:
A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.
<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

We are given:
Given mass of liquid bromine = 766. g
Molar mass of liquid bromine,
= 159.8 g/mol
Putting values in above equation, we get:

Hence, the amount of liquid bromine produced is 4.79 moles.
Answer:
1. All red calves i.e. RR
2. All roan calves i.e RW
3. 2 red calves (RR) and two roan calves (RW)
Explanation:
According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):
1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR
2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW
3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).
Answer:
1.403x10²⁴ molecules
Explanation:
In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first<u> convert grams to moles</u> using its <em>molar mass</em>:
- 102.5 g ÷ 44 g/mol = 2.330 mol CO₂
Now we <u>convert moles into molecules </u>using <em>Avogadro's number</em>:
- 2.330 mol * 6.023x10²³ molecules/mol = 1.403x10²⁴ molecules