Find the amount of heat (Q) needed to raise the temperature of 5.00 g of a substance from 20.0 C to 30.0C if

What is the relationship between Celsius and kelvin

Inga [223]

The relationship between Celsius and kelvin is that degrees Kelvin is equal to degrees Celsius + 273.15. That means that the freezing point of water (0°C) is 273.15° kelvin. This makes an additive relationship between the two quantities, so quantities such as the boiling point and freezing point of water are the same distance apart on the Kelvin scale (100 degrees). This means that kelvin has the same magnitude as Celsius.

4 0

2 years ago

Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate

Anvisha [2.4K]

Answer:

$-3272$ kJ/mol

Explanation:

Given and known facts

Mass of Benzene $= 0.187$ grams

Mass of water $= 250$ grams

Standard heat capacity of water $= 4.18$ J/g∙°C

Change in temperature ΔT $= 7.48$ °C

Heat

$=250 * 4.18 * 7.48\\=7816.6 \\=7.82$

Heat released by benzine is - 7.82 kJ

Now, we know that

$0.187$ grams of benzene release $= -7.82$ kJ heat

So, $1$ g benzine releases

$\frac{ -7.82 }{0.187}\\= -41.8$

kJ/g

$0.187 * \frac{1}{78.108}=0.00239$ mol C6H6

Heat released

$= \frac{-7.82}{ 0.00239}$

$=-3272$ kJ/mol

4 0

3 years ago

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

Ca: 40 g/mole
F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$