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Fiesta28 [93]
2 years ago
8

What happens when carbon dioxide gas is cooled to -78 degree Celsius​

Chemistry
1 answer:
katrin2010 [14]2 years ago
3 0

Answer:

At temperatures below −78 °C, carbon dioxide changes directly from a gas to a white solid called dry ice through a process called deposition.

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What best explains the type of energy present in the vibrating atoms of a substance
luda_lava [24]
The type of energy present in the vibrating atoms of a substance is a thermal energy and it is a kinetic type of energy. It is associated with movement within the crystal lattice of substance. ... Eventually, it can lead to motion of the atoms which is a form of kinetic energy.
7 0
3 years ago
How many moles of water contain 1.3 x 10^24 molecules?
Vika [28.1K]
C is the correct answer
3 0
4 years ago
Which of the following examples illustrates a number that is correctly rounded to three significant figures?
Naily [24]

Answer:

c. 20.0332 g to 20,0 g

Explanation:

A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.

<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures? </em>

a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.

b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures:  0.040.

c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures:  20.0.

d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.

e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.

8 0
3 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0
3 years ago
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