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Scorpion4ik [409]
3 years ago
14

How many moles in 28 grams of CO2?

Chemistry
1 answer:
Phantasy [73]3 years ago
4 0

# moles = mass / Mr

moles = 28 / (12+(16*2))

#moles = 28/44

#moles = 0.63 moles

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Grams of sodium 9.5g in NaCl <br>​
Kitty [74]

Answer:

3.68 grams.

Explanation:

First we <u>convert 9.5 g of NaCl into moles of NaCl</u>, using its<em> molar mass</em>:

9.5 g ÷ 58.44 g/mol = 0.16 mol NaCl

In<em> 0.16 moles of NaCl there are 0.16 moles of sodium </em>as well.

We now <u>convert 0.16 moles of sodium into grams</u>, using <em>sodium's molar mass</em>:

0.16 mol * 23 g/mol = 3.68 g

4 0
3 years ago
Which of the following isotopes is radioactive A)C-12 B)Pb-206 C) Tc-99 D)Ne-20
Gre4nikov [31]
C) Tc-99 is a radioactive isotope also known as radioisoptope. 
4 0
3 years ago
What happens to the pressure if the gas volume is cut in half while n and I are<br> held constant?
N76 [4]

Answer: You can use Boyle's law, which states that pressure is inversely related to volume when other variables are held constant. If the final pressure of a gas is half of the initial, the volume must double if temperature is to remain the same.

Explanation:

5 0
3 years ago
Select the correct terms to complete the statement:
Likurg_2 [28]

Answer:

The particles that make up a substance in its liquid state have <u>more </u>kinetic energy than those of the same substance in its solid-state.

For a solid to melt, energy must be <u>added to</u> the system.

For a liquid to freeze, energy must be <u>removed from</u> the system.

5 0
2 years ago
A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain
tatiyna

Answer:

C_m=0.474\frac{J}{g\°C}

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

Q_m=-Q_w

Thus, in terms of masses, specific heats and temperatures we can write:

m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}

Best regards!

6 0
2 years ago
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