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3 years ago

What are thermal plasmas?

2 answers:

6 0

<span>plasmas that reach a temperature equal to their surroundings

----------------------=0uyjgyhjngytfgbhtfgbdc b gdf gb </span>

Evgesh-ka [11]3 years ago

5 0

In short the answer is b

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A titration of 0.1 M NaOH into 1.2 L of HCl was stopped once the pH reached 7 (at 25C). If 0.4 L of NaOH needed to be added to a

MArishka [77]

Answer:

0.033 M

Explanation:

Let's consider the neutralization reaction between NaOH and HCl.

NaOH + HCl → NaCl + H₂O

0.4 L of 0.1 M NaOH were used. The reacting moles of NaOH are:

0.4 L × 0.1 mol/L = 0.04 mol

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 0.04 moles.

0.04 moles of HCl are in 1.2 L. The molarity of HCl is:

M = 0.04 mol / 1.2 L = 0.033 M

4 0

3 years ago

20 points! Help ASP! Your answer must be explained!

Mariana [72]

<span>b.mixtures are made up of more than one component</span>

5 0

3 years ago

Read 2 more answers

What is NOT a property of a metal?

Phantasy [73]

Answer:

Lustrous is NOT a property of metal

6 0

3 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

3 years ago

1.this substance is a natural non-living material with a uniform structure throughout

vovikov84 [41]

1.Mineral
2.Igneous
3.Composition

7 0

4 years ago

Read 2 more answers