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spin [16.1K]
3 years ago
6

Give one example of a Phase Change

Chemistry
1 answer:
Tom [10]3 years ago
4 0
Ice melting into water
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Scientists are studying different jackets made for extremely cold, snowy weather. They are trying to find jackets
koban [17]

Answer:

Which jacket traps the most heat.

Explanation:

I took the test

8 0
3 years ago
The. bond dissociation enthalpies of the H-H bond and the H-Cl bond are 435 kJ mol^-1 and 431 kJ mol^-1, respectively. The ΔHfO
Novay_Z [31]

The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

H-H bond energy =  435 kJ mol^-1

H-Cl bond energy = 431 kJ mol^-1

ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

-92 = 435  +  431 + x

x = -92 - (435  +  431)

x = -958 kJ mol^-1

Learn More about dissociation enthalpy:brainly.com/question/9998007?

#SPJ1

6 0
2 years ago
A sample of gas occupies 9.0 mL at a pressure of 500.0 mm Hg. A new volume of the same sample is at a pressure of 750.0 mm Hg.
Anuta_ua [19.1K]
<h3>Answer:</h3>

                The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

<h3>Solution:</h3>

              According to Boyle's Law, " <em>The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure</em>". Mathematically, the initial and final states of gas are given as,

                                     P₁ V₁  =  P₂ V₂    ----------- (1)

Data Given;

                  P₁  =  500 mmHg

                  V₁  =  9.0 mL

                  P₂  =  750 mmHg

                  V₂  =  ??

Solving equation 1 for V₂,

                   V₂  =  P₁ V₁ / P₂

Putting values,

                   V₂  =  (500 mmHg × 9.0 mL) ÷ 750 mmHg

                   V₂  =  6.0 mL

<h3>Result:</h3>

            The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

4 0
3 years ago
Read 2 more answers
pyrite is a mineral composed of 46.5 mass % iron and 53.5 mass % sulfur. determine the empirical formula for calcocite.
padilas [110]

The empirical formula for pyrite is FeS2.

HOW TO CALCULATE EMPIRICAL FORMULA:

  • The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:

46.5 mass % Fe = 46.5g of Fe

53.5 mass % S = 53.5g of S

  • Next, we divide each element's mass value by its molar mass

Fe = 46.5g ÷ 56g/mol = 0.83mol

S = 53.5g ÷ 32g/mol = 1.67mol

  • Next, we divide each mole value by the smallest (0.83mol)

Fe = 0.83mol ÷ 0.83 = 1

S = 1.67mol ÷ 0.83 = 2.014

Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.

Learn more at: brainly.com/question/14044066?referrer=searchResults

3 0
2 years ago
If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t
tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of CaCO_3

\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}

Given:

Mass of CaCO_3 = 0.524 g

Molar mass of CaCO_3 = 100 g/mol

\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol

Now we have to calculate the concentration of CaCO_3

\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}

So,

Concentration of calcium ion = Concentration of CaCO_3 = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

4 0
3 years ago
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