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Molodets [167]
3 years ago
6

5. How are the members of Kingdom Archaebacteria different from the

Chemistry
1 answer:
yan [13]3 years ago
5 0

Answer:

5. C

6. B

Explanation:

sana nakatulong

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Write the balanced chemical equation where liquid decane burns in oxygen gas to form carbon dioxide gas and water vapor. Use the
Allushta [10]

Answer:

2C_{10} H_{22(l)} + 31O2(g)--> 20CO_{2(g)} + 22H_{2}O_{(g)}

Explanation:

2C_{10} H_{22(g)} + 31O2(g)--> 20CO_{2(g)} + 22H_{2}O_{(g)}

From the question, one can work out which states of matter to assign to which species. The trick with organic equations of this nature is to try to balance everything but oxygen first. Make sure you balance oxygen last because it is the easiest to balance.

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How do disease causing organism reach from one organism to other​
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Infectious diseases are commonly transmitted through direct person-to-person contact.

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4 0
3 years ago
What is another common name for the lanthanoids on the periodic table?
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Lanthanide Lanthanoid, also called Lanthanide
7 0
3 years ago
The reaction described in Part A required 3.54 L of magnesium chloride. What is the concentration of this magnesium chloride sol
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7 0
3 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0
3 years ago
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