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natta225 [31]
3 years ago
10

PLEASE HELP I WILL GIVE BRAINLIEST

Chemistry
1 answer:
stellarik [79]3 years ago
3 0

Answer: YES!

Explanation:

Ox2 Co2 Hydrogen and Corossion 3

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A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C^−1 , is put into a calorimeter (see sketch at right) t
Tcecarenko [31]

Answer: The mass of aluminium sample is 55.4 gram

Explanation:

Q_{absorbed}=Q_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      

where,

m_1 = mass of aluminium = ?

m_2 = mass of water = 300.0 g

T_{final} = final temperature = 23.8^0C

T_1 = temperature of aluminium = 94.5^oC

T_2 = temperature of water = 21.0^oC

c_1 = specific heat of aluminium = 0.897J/g^0C

c_2 = specific heat of water =  4.184J/g^0C

Now put all the given values in equation (1), we get

-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)]

-[m_1\times 0.897\times (23.8-94.5)^0C]=[300.0g\times 4.184\times (23.8-21.0)]

m_1=55.4g

Therefore, the mass of aluminium sample is 55.4 gram

8 0
3 years ago
Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
Katen [24]

Answer:

We will produce 57.1 grams of silver sulfadiazine

Explanation:

Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If 20.0 g of Ag2O is reacted with 40.0 of C10H10N4SO2, what mass of silver sulfadiazine (AgC10H9N4SO2) can be produced assuming 100% yield?

Step 1: Data given

Mass of Ag2O = 20.0 grams

Molar mass of Ag2O = 231.735 g/mol

Mass of C10H10N4SO2 = 40.0 grams

Molar mass of C10H10N4SO2 = 250.277 g/mol

Step 2: The balanced equation

Ag2O + 2C10H10N4SO2 → 2AgC10H9N4SO2 + H2O

Step 3: Calculate moles Ag2O

Moles Ag2O = mass Ag2O / molar mass Ag2O

Moles Ag2O = 20.0 grams / 231.735 g/mol

Moles Ag2O = 0.0863 moles

Step 4: Calculate moles C10H10N4SO2

Moles C10H10N4SO2 = 40.0 grams / 250.277 g/mol

Moles C10H10N4SO2 = 0.160 moles

Step 5: Calculate the limiting reactant

For 1 mol Ag2O we need 2 moles 2C10H10N4SO2 to produce 2 moles AgC10H9N4SO2 and 1 mol H2O

C10H10N4SO2 is the limiting reactant. There will react 0.160 moles.

Ag2O is in excess. There will react 0.160/ 2 = 0.080 moles

There will remain 0.0863 - 0.080 = 0.0063 moles Ag2O

Step 5: Calculate moles AgC10H9N4SO2

For 1 mol Ag2O we need 2 moles 2C10H10N4SO2 to produce 2 moles AgC10H9N4SO2 and 1 mol H2O

For 0.160 moles C10H10N4SO2 we'll have 0.160 moles AgC10H9N4SO2

Step 6: Calculate mass AgC10H9N4SO2

Mass AgC10H9N4SO2 = moles * molar mass

Mass AgC10H9N4SO2 = 0.160 moles * 357.137 g/mol

Mass AgC10H9N4SO2 = 57.1 grams

We will produce 57.1 grams of silver sulfadiazine

6 0
3 years ago
Please show work, im really confused so please help!!
Rama09 [41]
No because the mass of the empty beaker , mass of beaker  is not balanced the amount of alcohol in the beaker.
4 0
3 years ago
Light strikes a beautiful white perch under water. This light is reflected back to your eyes allowing you to see the fish. As th
Ivahew [28]
The answer is: <span>Light strikes a beautiful white perch under water. This light is reflected back to your eyes allowing you to see the fish. As the light left the water it changes speed causing the light to bend away from the normal.
Hope this helps!
(i got this from </span>https://quizlet.com/5474123/chapter-13-light-flash-cards/ if you need more help)
7 0
3 years ago
An ethylene gas torch requires 300 L of gas at 0.8 atm. What will be the pressure of the gas if ethylene is supplied by a 200.0
Vlad [161]

Answer:

1.2 atm

Explanation:

Given data

  • Volume of the gas in the tank (V₁): 200.0 L
  • Pressure of ethylene gas in the tank (P₁): ?
  • Volume of the gas in the torch (V₂): 300 L
  • Pressure of the gas in the torch (P₂): 0.8 atm

If we consider ethylene gas to be an ideal gas, we can find the pressure of ethylene gas in the tank using Boyle's law.

P_1 \times V_1 = P_2 \times V_2\\P_1 = \frac{P_2 \times V_2}{V_1} = \frac{0.8atm \times 300L}{200.0L} = 1.2 atm

3 0
3 years ago
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