Answer:
Mass of sea food = 30.98 Kg
Mass of sea food in pound = 68.31 lbs
Explanation:
Salmon, crab and oysters all are sea food.
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
Mass of salmon = 22 kg
Mass of crab = 5.5 kg
Mass of oysters = 3.48 kg
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
= 22 + 5.5 + 3.48
= 30.98 Kg
1 Kg = 2.205 lbs
Therefore, 30.98 kg = 30.98 × 2.205
= 68.31 lbs
The change in temperature of the metal is 6.1°C. Details about change in temperature can be found below.
<h3>How to calculate change in temperature?</h3>
The change in temperature of a substance can be calculated by subtracting the initial temperature of the substance from the final temperature.
According to this question, a 25.0 g sample of metal at 16.0 °C is warmed to 22.1 °C by 259J of energy.
This means that the change in temperature of the metal can be calculated as:
∆T = 22.1°C - 16°C
∆T = 6.1°C
Therefore, the change in temperature of the metal is 6.1°C.
Learn more about change in temperature at: brainly.com/question/19051558
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<h2>
Answer: 131.9 g</h2>
<h3>
Explanation:</h3>
<u>Write a Balanced Equation for the decomposition</u>
CaCO₃ → CaO + CO₂
<u></u>
<u>Find Moles of CO₂ Produced</u>
Since the mole ratio of CaCO₃ to CO₂ is 1 to 1,
the moles of CaCO₃ = moles of CO₂
moles of CaCO₃ = mass ÷ molar mass
= 300 g ÷ 100.087 g/mol
= 2.997 moles
∴ moles of CO₂ = 2.997 moles
<u>Determine Mass of CO₂</u>
Mass = moles × molar mass
= 2.997 mol × 44.01 g/mol
= 131.9 g
<u></u>
<h3>∴ when 300 g of calcium carbonate is decomposed, it produces 131.9 g of carbon dioxide.</h3>
2Agl + 1Na2S----->1 Ag2S + 2Nal
This may help
Answer:
Scientific inquiry
Explanation:
A scientific inquiry is a systematic way of finding out about certain phenomena in our environment through the use of careful observations and experimentation. It follows a series of steps and procedures which helps a scientist to critically think and observe certain events in his environment. Through the observations, scientific conclusions are then drawn.