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2 years ago

Glass has a specific heat of 0.16 calories/gram°C.

2 answers:

7 0

Calories heat added = Mass of glass times temp. increase times specific heat of glass

calories (small calories) = l.0 g x 20 degrees x .16 calories/gm/degree C = 3.2 calories

nignag [31]2 years ago

4 0

Answer:

3.2 calories

Explanation:

The state of aggregation of the glass does not change, as its temperature increases, it is a sensible heat, because it depends on its temperature.

The sensible heat formula is

$Q= m.sh.T$

$sh= 0.16\frac{cal}{g C}$

$m= 1.0 g\\T= 20C$

$Q= 1.0 g.0.16\frac{cal}{gC} .20C\\ Q= 3.2 cal$

1.0 gram of glass will rise 20.0°C when 3.2 calories of heat are applied

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A sample of carbon dioxide is contained in a 250.0 mL flask at 0.930 atm and 15.4 °C. How many molecules of gas are in

Yanka [14]

Answer:

We are given:

Volume (V) = 0.25 L

Pressure (P) = 0.93 atm

Temperature (T) = 15.4°C OR 288.4 K

Solving for the number of moles of CO₂:

From the ideal gas equation:

PV = nRT

replacing the variables

0.93 * 0.25 = n (0.082)(288.4)

n = 0.00983 moles

Number of molecules:

Number of moles= 0.00983

number of molecules in 1 mole = 6.022 * 10²³

Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³

Number of molecules = 5.91 * 10²¹

8 0

2 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

2 years ago

2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in

poizon [28]

Answer: $3.8\times 10^{-3}M$

Explanation:

Moles of $NOCl$ = 1 mole

Moles of $Cl_2$ = 1 mole

Volume of solution = 1 L

Initial concentration of $NOCl$ = 1 M

Initial concentration of $Cl_2$ = 1 M

The given balanced equilibrium reaction is,

$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)$

Initial conc. 1 M 0M 1 M

At eqm. conc. (1-2x) M (2x) M (1+x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}$

The $K_c$ = $1.6\times 10^{-5}}$

Now put all the given values in this expression, we get :

${1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}$

By solving the term 'x', we get :

$x=0.0019$

Concentration of $NO$ at equilibrium= (2x) M = $2\times 0.0019=3.8\times 10^{-3}M$

6 0

2 years ago

Consider four elements from Group 7A: fluorine in the second period, chlorine in the third period, bromine in the fourth period,

lina2011 [118]

Out of the possible answers for this question, fluorine in the second period is correct. Of the four elements fluorine, chlorine, bromine and iodine, fluorine has the largest first ionization energy, with a Enthalpy number of 1681.0. Of all the elements, helium has the highest first ionization energy figure.

5 0

3 years ago

Is calcium carbonate a compound element or mixture?

uysha [10]

It's a compound because it's formed from an ionic bond between calcium and carbon so it's an ionic compound

7 0

3 years ago