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otez555 [7]
3 years ago
11

Help plzzzzzz ill give brainliest​

Chemistry
1 answer:
Stells [14]3 years ago
5 0
The answer is the first bubble, it absorbs white light and refracts green, this is because the color you see is the reflection of the object
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How is a line spectra created from a hot gas? a. photons are emitted when the amount of radioactive material in the gas decrease
fiasKO [112]

Answer:

c when electrons drop down from a high energy level they emit a photon.

Explanation:

4 0
1 year ago
Could the Periodic Table be arranged differently?
coldgirl [10]

Answer:

No because it is stayed that way and you can't define them differently.

7 0
3 years ago
Read 2 more answers
The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma
Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C=p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water=n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol

Moles of ethylene glycol=n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0
3 years ago
A student wants to make a 0.600 M aqueous solution of barium sulfate, BaSO4, and has a bottle containing 12.00 g of barium sulfa
love history [14]

<u>Answer: </u>The volume of the solution is 85.7 mL

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}} .....(1)

We are given:

Molarity of solution = 0.600 M

Given mass of BaSO_4 = 12.00 g

We know, molar mass of BaSO_4=[(1\times 137.33)+(1\times 32.07)+(4\times 16)]=233.4g/mol

Putting values in equation 1, we get:

0.600=\frac{12.00\times 1000}{233.4\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{12.00\times 1000}{233.4\times 0.600}=85.68mL=85.7mL

The rule of significant number that is applied for the problems having multiplication and division:

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

Here, the least number of significant figures is 3 that is determined by the number, 0.600. Thus, the answer must have these many significant figures only.

Hence, the volume of the solution is 85.7 mL

5 0
3 years ago
The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co
garri49 [273]

Answer:

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater (m_{sw}), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

m_{sw} = \rho_{sw}\cdot V_{sw}

Where:

\rho_{sw} - Density of seawater, measured in grams per liters.

V_{sw} - Volume of seawater, measured in liters.

If V_{sw} = 1.5\times 10^{21}\,L and \rho_{sw} = 1030\,\frac{g}{L}, then:

m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)

m_{sw} = 1.545\times 10^{24}\,g

The total mass of sodium chloride is determined by the following ratio:

r = \frac{m_{NaCl}}{m_{sw}}

m_{NaCl} = r\cdot m_{sw}

Given that m_{sw} = 1.545\times 10^{24}\,g and r = 0.035, the total mass of sodium chloride in all the seawater in the world is:

m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)

m_{NaCl} = 5.408\times 10^{22}\,g

There are 5.408\times 10^{22} grams contained in all the seawater in the world.

8 0
3 years ago
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