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iogann1982 [59]
2 years ago
15

Sigh.....................................

Chemistry
2 answers:
attashe74 [19]2 years ago
7 0

Answer:

That is a nice picture i love it plus if you need help just ask we can help :3

Explanation:

3241004551 [841]2 years ago
5 0

Answer:

what wrong

Explanation:

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135-g sample of a metal requires 2.50 kJ to change its temperature from 19.5 C to 100.0 C what is the specific heat of this meta
Harman [31]
The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5. 
3 0
3 years ago
Which best describes the relationship between Haiti’s political instability and the natural disasters it has faced?
Paul [167]

Explanation:

FOR ALL MY POINTS write an argumentative essay on why dog parks are "helpful or harmful"

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8 0
2 years ago
What is the mass of substance that exactly fills a 247.2 mL container? The density of the substance is 0.81 g/mL. Give the answe
matrenka [14]

Answer:

<h2>mass = 200.23 g</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

Since we are finding the mass

<h3>mass = Density × volume</h3>

From the question

Density = 0.81 g/mL

volume = 247.2 mL

Substitute the values into the above formula and solve for the mass

mass = 0.81 × 247.2

= 200.232

We have the final answer as

<h3>mass = 200.23 g to 2 decimal places</h3>

Hope this helps you

3 0
3 years ago
Suppose you have a sample of table salt (NaCl) that has a mass of 100.0 grams. If
IgorC [24]

Taking into account the definition of molarity, the concentration of the solution is 0.855 \frac{moles}{liter}.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

Molarity=\frac{number of moles}{volume}

Molarity is expressed in units \frac{moles}{liter}.

<h3>Molarity of NaCl</h3>

In this case, you have:

  • number of moles of NaCl= 100 grams\frac{1 mole}{58.45 grams} =1.71 moles (being 58.45 g/mole the molar mass of NaCl)
  • volume 2 L

Replacing in the definition of molarity:

Molarity=\frac{1.71 moles}{2 L}

Solving:

Molarity= 0.855 \frac{moles}{liter}

Finally, the concentration of the solution is 0.855 \frac{moles}{liter}.

Learn more about molarity:

<u>brainly.com/question/9324116</u>

<u>brainly.com/question/10608366</u>

<u>brainly.com/question/7429224</u>

5 0
2 years ago
"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

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

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0
3 years ago
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