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natka813 [3]
3 years ago
7

HELP ASAP!!!

Chemistry
1 answer:
sammy [17]3 years ago
8 0

Answer:The answer is number 4 aka 7

Explanation:

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viva [34]
4 significant figures
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3 years ago
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A standard gold bar has a mass of 12.4 kg. When the gold bar is placed into a container of water the volume changes by 642 mL. W
AlekseyPX

Answer:

19.3 g/mL

Explanation:

The following data were obtained from the question:

Mass (m) = 12.4 kg

Volume (V) = 642 mL.

Density (D) =.?

Next, we shall convert 12.4 kg to grams. This can be obtained as follow:

1 kg = 1000 g

Therefore,

12.4 kg = 12.4 × 1000

12.4 kg = 12400 g

Therefore, 12.4 kg is equivalent to 12400 g.

Finally, we shall determine the density of the gold as follow:

Density is simply defined as the mass of the substance per unit volume of the substance. It can be represented mathematically as:

Density (D) = mass (m) / volume (V)

D = m/V

With the above formula, the density of gold can easily be obtained as follow:

Mass (m) = 12400 g

Volume (V) = 642 mL.

Density (D) =.?

D = m/V

D = 12400/642

D = 19.3 g/mL

Therefore, the density of hold is 19.3 g/mL

6 0
3 years ago
How many photons are contained in a burst of yellow light (589 nm) from a sodium lamp that contains 623 kJ of energy?
Goryan [66]
<span>First we need to find the energy of one photon with a wavelength of 589 nm. E = hc / wavelength E = (6.63 x 10^{-34} J s)(3 x 10^8 m/s) / (589 x 10^{-9} m) E = 3.3769 x 10^{-19} Joules To find N, the number of photons, we need to divide the total energy by the energy of each photon. N = 623000 J / 3.3769 x 10^{-19} Joules N = 1.84 x 10^{24} photons There are 1.84 x 10^{24} photons in the burst of yellow light.</span>
3 0
3 years ago
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Calculate the vapor pressure of spherical water droplets of radius (a) 17 nm and (b) 2.0 μm surrounded by water vapor at 298 K.
Montano1993 [528]

Explanation:

Relation between pressure of water and its droplet is as follows.

           ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}

where,   p = pressure of droplet

          p_{o} = water pressure in given temperature

          \gamma = 7.99 \times 10^{-3}

           M = Molecular Weight in Kg/Mol (0.018 for water)

            r = radius in meters

     \rho = density of water in Kg/m^{3} (1000 kg/m^{3})

           R = ideal gas constant (8.31)

           T = temperature in Kelvin

(a)   We will calculate the value of p as follows.

           p = e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}

              = e^{\frac{2 \times 0.07199 \times 0.018}{1.7 \times 10^{-8} \times 1000 \times 8.31 \times 298 K} \times 25.2

              = 26.8 torr

(b)  And, vapor pressure of spherical water droplets of radius 2.0 \mu m or 2 \times 10^{-6} m

             p = e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}

              = e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2

              = 25.2 torr

7 0
4 years ago
Every ribosome has three tRNA binding sites: the A site, the P site, and the E site. For the given sentences, select the correct
alex41 [277]

Answer:

A. All sites B. A site C. E site D. P site

Explanation:

A. Because all sites consequently take place while peptide is synthesized.

B. A site (aminoacyl site) is called so because it binds to charged aminoacyl peptide.

C. Because it is the exit site and it takes the final stage on the uncharged molecule.

D. The P site (peptidil site which takes main role in peptide synthesis.

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3 years ago
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