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zhuklara [117]
2 years ago
12

How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K

Chemistry
1 answer:
Olenka [21]2 years ago
6 0

I’m not sure I need help with this question

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The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.
xxMikexx [17]

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

6 0
2 years ago
What is the mass of a sample of water containing 3.55×1022 molecules of h2o?
ad-work [718]
6,02×10²³ -------------- 18g
3,55×10²² --------------  xx=\frac{3,55*10^{22}*18g}{6,02*10^{23}}=10,61*10^{22-23}g=10,61*10^{-1}g=1,061g


6 0
2 years ago
Read 2 more answers
A gram of gasoline produces 45.0kj of energy when burned. gasoline has a density of 0.77/gml . how would you calculate the amoun
babymother [125]
Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:

48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ

<span>The amount of energy produced by burning 48 L of gasoline was then determined to be 1663200 kJ. </span>    
5 0
2 years ago
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°C _____.
MatroZZZ [7]

Answer:

c

Explanation:

1 calorie = 4.184J/g×°C

This also happens to be the specific heat capacity of water, which is the amount of energy it takes to raise the temperature of 1mL of water by 1°C

6 0
2 years ago
Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p
Solnce55 [7]

Answer:

1.17 grams

Explanation:

Let's consider the balanced equation for the combustion of ethylene.

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

  • 1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
  • The molar mass of C₂H₄ is 28.05 g/mol.

The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}

8 0
2 years ago
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