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3 years ago

3.25 x 10+8 nm2 divide by 6.5 x 10+6 nm =

Chemistry

1 answer:

Vesnalui [34]3 years ago

8 0

Answer: 50 nm

Explanation: Two steps:

1. Divide 3.25/6.5 = 0.5

2. Divide 10^8/10^6 = 10^2

nm^2/nm = nm

Combine: 0.5x10^2 nm

or 50 nm

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A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi

Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = 0.0348g of water

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = 9.9652g of water

I hope it helps!

4 0

3 years ago

A gas has an initial volume of 455 mL at 105ºC and a final volume of 235 mL. What is its final temperature in Celsius degrees?

Oksana_A [137]

Hello!

To solve this problem we're going to use the Charles' Law. This Law describes the relationship between Volume and Temperature in an ideal gas. Applying this law we have the following equation:

$\frac{V1}{T1} = \frac{V2}{T2} \\ \\ T2= \frac{V2*T1}{V1}= \frac{235 mL * 105 ^{\circ}C }{455 mL}=54,23 ^{\circ}C$

So, the final temperature is 54,23 °C

Have a nice day!

5 0

4 years ago

Read 2 more answers

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

Calculate the density of carbon dioxide at STP

Artist 52 [7]

Answer:

Density = mass/volume

= 44/22.4

= 1.96 gram/liter

The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.

5 0

3 years ago

What are two things that you observed when you heated the mixture of tin and nitric acid over the Bunsen burner in the virtual l

Rus_ich [418]

1. The reaction for this would be:

Sn + 4 HNO₃ → SnO₂ + 4 NO₂ + 2 H₂O

The first observation would be bubbling of the solution and brown acrid smoke is produced due to the presence of NO₂ gas. Another observation would be the presence of a white solid which is SnO₂.

2. Heating was required to get rid of the H₂O. When all moisture is gone, you weigh the sample. Afterwhich, you further heat it to get ride of the oxygen. By doing this, you would know the individual mass of each element. Then, you can solve for the empirical formula of the oxide of tin.

3 0

4 years ago