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Answer:
1) CO₂
2) 0.2551 g
Explanation:
The balanced reactions are:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂
1) The gas produced is CO₂.
2) Calculate mass of CaCO₃:
(0.5236 g) (0.4230) = 0.2215 g CaCO₃
Convert to moles:
(0.2215 g CaCO₃) (1 mol / 100.1 g) = 0.002213 mol CaCO₃
Find moles of CaCO₃:
(0.002213 mol CaCO₃) (1 mol CO₂ / mol CaCO₃) = 0.002213 mol CO₂
Convert to mass:
(0.002213 mol CO₂) (44.01 g / mol) = 0.09738 g CO₂
Calculate mass of MgCO₃:
(0.5236 g) (0.5770) = 0.3021 g MgCO₃
Convert to moles:
(0.3021 g MgCO₃) (1 mol / 84.31 g) = 0.003583 mol MgCO₃
Find moles of MgCO₃:
(0.003583 mol MgCO₃) (1 mol CO₂ / mol MgCO₃) = 0.003583 mol CO₂
Convert to mass:
(0.003583 mol CO₂) (44.01 g / mol) = 0.1577 g CO₂
Total mass of CO₂:
0.09738 g CO₂ + 0.1577 g CO₂ = 0.2551 g CO₂
ANSWER FOR QUESTION 1:
The most common type of E. coli infection that causes illness in people is called E. coli O157, which produces a toxin known as Shiga-toxin. Shiga-toxin producing E. coli is abbreviated as STEC. Symptoms of infection with this germ include watery or bloody diarrhea, fever, abdominal cramps, nausea, and vomiting.
ANSWER FOR QUESTION 2:
Some other germs don’t cause as many illnesses, but when they do, the illnesses are more likely to lead to hospitalization. Those germs include: Anyone can get sick from eating contaminated food. Follow four simple food safety steps —clean, separate, cook, and chill—to lower your chance of food poisoning and to protect yourself and your loved ones.
Answer:
The disease is autosomal dominant.
Explanation:
Huntigton's disease is an autosomal dominant disease because the allele for this disease is present on an autosomal chromosome and the person with even one mutated allele (H) can develop the disease even if he has one normal allele (h) too. If a person is affected with Huntigton's disease, there are 50 percent chances that the children will also suffer from the disease.
For example: A father is suffering from Huntigton's disease but mother is normal. Let us see how it will be passed to kids.
P1: Hh : hh
Gametes: H : h: h: h
Offspring: Hh: Hh: hh: hh
50% : 50%
Therefore, 50 percent chances are there that the kids will have disease even if only parent suffers from it.