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In-s [12.5K]
3 years ago
12

why are plasmas of great interest to scientists or manufacturers. descride 2 current uses of plasmas and describe 1 way scientis

ts and engineers hope to use plasmas in the future
Chemistry
1 answer:
Helga [31]3 years ago
6 0

<u>Answer:</u>

<u>Plasmas of great interest to scientists or manufacturers as</u>

  • Plasma is electrically charged gases that contain considerable charged particles that can change the behavior of the substance.

<u>Current uses of plasmas:</u>

  • First, it is used to make semiconductors for different types of electronic equipment
  • Secondly, they're used in making transmitters for high-temperature films.

<u>Way scientists and engineers hope to use plasmas in the future:</u>

  • The scientists are hoping to use plasma in the future to get rid of all hazardous wastes through a process called plasma gasification.

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Which of the following may indicate that a chemical reaction has occurred?
lianna [129]

Answer:

D, all of the above

Explanation:

3 0
2 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
2 years ago
What is the molarity of 50.84 g of Na2CO3 dissolved in 0.400 L solution?
borishaifa [10]

Answer:

1.20 M

Explanation:

Convert grams of Na₂CO₃ to moles.  (50.84 g)/(105.99 g/mol) = 0.4797 mol

Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M

4 0
3 years ago
In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical
Umnica [9.8K]

Grams of Phosphorus = 4.14 grams 
Grams of white compound = 27.8 grams 
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
 Calculating moles which would be grams / molar mass
 Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
 Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
 Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
 Calculating the ratios by dividing with the small entity
 P = 0.1337 moles / 0.1337 moles = 1
 Cl = 0.6674 moles / 0.1337 moles = 5 
So the empirical formula would be PCl5
3 0
3 years ago
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
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