A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the cabinet is used to store bottles of pentane, tetrahydrofuran, carbon tetrachloride, methyl acetate, and ethanolamine. The chemist plans to try to identify the unknown liquid by measuring the density and comparing to known densities. First, from his collection of Material Safety Data Sheets (MSDS), the chemist finds the following information: liquid density pentane tetrahydrofuran carbon tetrachloride methyl acetate ethanolamine Next, the chemist measures the volume of the unknown liquid as and the mass of the unknown liquid as . Calculate the density of the liquid. Round your answer to significant digits. Given the data above, is it possible to identify the liquid

Answer:

3 ................................

Answer: 3

Explanation:

yes

Answer:

Synthesis

Explanation:

A sysnthesis reaction takes two or more reactants to create one product

Answer:

C is the only reasonable answer.. but this is 6th grade science and I'm in 7th, so I'm pretty sure I'm right

Answer:

97.95 g/mol

Explanation:

Step 1: Define molar masses

Molar Mass of N - 14.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of Br - 79.90 g/mol

Step 2: Find amounts of molar masses

1 mol N - 14.01 g/mol

4 mol H - 4.04 g/mol

1 mol Br - 79.90 g/mol

Step 3: Find molar mass of compound

Molar Mass N + 4 Molar Mass H + 1 Molar Mass Br

14.01 g/mol + 4.04 g/mol + 79.90 g/mol

97.95 g/mol

Answer:

Either 12:12:6 or 6:6:12

Explanation:

If I remember correctly the answer should be 6:6:12

Answer:

1.186L or 1186mL.

Explanation:

The volume of a 0.290 m NH4I solution that is required to react with 253 ml of a 0.680 m Pb(NO3)2 solution is 1.186 liters or 1186 ml.

Answer:

Metals do not form double and triple bonds in general because metals want to lose electrons, not share, in order to become stable, and both compounds have high melting ppints

Compounds with metallic bonds and ionic compounds are similar in their melting point and boiling point, conductivity, solubility in polar solvents etc.

Ionic compounds are chemical compounds that are formed by the electrostatic attraction between oppositely charged ions.

Compounds with metallic bonds and ionic compounds are similar in several ways. Some of the similarities are:

1. High melting and boiling points: Both metallic and ionic compounds have high melting and boiling points due to the strong forces that hold the atoms or ions together.

2. Conductivity: Both metallic and ionic compounds are good conductors of electricity due to the presence of charged particles that can move freely.

3. Brittle: Ionic and metallic compounds are generally brittle in nature, meaning that they are prone to breaking or shattering when subjected to stress.

4. Solubility in polar solvents: Both metallic and ionic compounds are generally soluble in polar solvents such as water, due to the polar nature of the molecules.

In conclusion, compounds with metallic bonds and ionic compounds share several similarities. Both have high melting and boiling points, are good conductors of electricity, are brittle, and are generally soluble in polar solvents.

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Answer:

Hydrogen sulfide is an newly recognized signical molecule with very potent cytoprotective actions.

Answer:

Explanation:

A proton has a mass of approximately 938 MeV/c2, of which the rest mass of its three valence quarks contributes only about 9.4 MeV/c2; much of the remainder can be attributed to the gluons' QCBE.

A small pebble sinks in water because its density is greater than the density of water. Density is a measure of how much mass is packed into a given volume. If an object is denser than the liquid it is placed in, it will sink.

When a pebble is dropped into water, the water exerts an upward force called buoyancy. This force opposes the weight of the pebble, and if the pebble is less dense than water, it will experience an upward force greater than its weight and float.

However, if the pebble's density is greater than that of water, the downward force due to its weight will be greater than the upward buoyant force, causing it to sink.

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Answer:

Theoretical yield = 0.1 g

Percent yield = 80%

Explanation:

Given data:

Mass of sodium = 2.3 g

Actual yield of hydrogen = 0.080 g

Theoretical yield = ?

Percent yield = ?

Solution:

Chemical equation:

2Na + 2H₂O    →  H₂ + 2NaOH

Number of moles of sodium:

Number of moles = mass/ molar mass

Number of moles = 2.3 g/ 23 g/mol

Number of moles = 0.1 mol

Now we will compare the moles of sodium with hydrogen to calculate the theoretical yield.

                       Na               :              H₂

                        2                :               1

                     0.1                 :             1/2×0.1 = 0.05 mol

Mass of hydrogen produced:

Mass = number of moles × molar mass

Mass = 0.05 mol × 2 g/mol

Mass = 0.1 g

Percent yield of hydrogen:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 0.08 g/ 0.1 g×100

Percent yield = 0.8×100

Percent yield = 80%

Answer:

Amplitude

Explanation:

Answer:

The rate of reaction remains the same, no change is observed

Explanation:

Remember that for an SN2 reaction, the rate of reaction depends both on the concentration of the alkyl halide and the concentration of the nucleophile.

Hence we can write; Rate = k [Alkyl halide][NaI]

This implies that if we half the concentration of the nucleophile and double the concentration of the alkyl halide, the rate of reaction just remains the same since the reaction is bimolecular and first order in both alkyl halide and nucleophile

Answer:

Explanation:

the SI unit of force. It is equal to the force that would give a mass of one kilogram an acceleration of one meter per second per second, and is equivalent to 100,000 dynes.

Answer:

I,II, III

Explanation:

Firstly, the magnitude of nuclear charge affects the first ionization energy of an element. Hence,as effective nuclear charge increases, the attraction between the nucleus and the outermost electron increases and ionization energy consequently increases. Thus Mg has a higher first ionization energy than Be.

Secondly, oxygen has an electron configuration of 1s2 2s2 2p4 while nitrogen has an electron configuration of 1s2 2s2 2p3. Now, recall that extra energy is often associated with half filled orbitals hence nitrogen has a higher first ionization energy than oxygen. Furthermore, the addition of electron to an already half filled 2p orbital in oxygen (pairing) leads to inter electronic repulsion and drastic fall in first ionization energy. Therefore, as we move from nitrogen to oxygen in the periodic table, greater inter-electron repulsion between two electrons in the same p-orbital counter balances the increase in effective nuclear charge hence nitrogen has a greater first ionization energy than oxygen.

Lastly, the 3p orbital is far away from the nucleus hence we expect it to feel less of nuclear attraction than a 2p orbital. Hence the first ionization energy of Ar is less than that of Ne.

Answer:

It is generally arranged by the atomic number

Explanation:

Answer:

The law of superposition states that each rock layer is older than the one above it.  So, the relative age of the rock or fossil in the rock is older if it is farther down in the rock layers.

Answer:

C)1 and 3 only☺️

Explanation:

Answer:

K

Explanation:

Answer:

The rest of the food molecule is converted to other carbon molecules. The hawk cannot consume all the parts of the squirrel. It will feed as a carnivore on the fleshy part of its prey. The matter is converted to other carbon molecules

Explanation:

Answer:

hi :D

Explanation:

State the relationship between temperature and kinetic energy. The Kinetic Molecular Theory allows us to explain the existence of the three phases of matter: solid, liquid, and gas. In addition, it helps explain the physical characteristics of each phase and how phases change from one to another.

hope this helps

._.

...

Answer:

Watts are a measurement of power, describing the rate at which electricity is being used at a specific moment. For example, a 15-watt LED light bulb draws 15 watts of electricity at any moment when turned on. Watt-hours are a measurement of energy, describing the total amount of electricity used over time.

Please mark as brainliest.

Answer: True

Watts are a measurement of power, used in light bulbs and other electricity. It is also measured as candela (C) in SI Units. The answer to the question is true because of this measurement.

Hope this helps!

Answer:

chongus because he's the only good one

Answer:

2.322 half-lives have passed to decay down from 150 miligrams to 30 miligrams.

Explanation:

The half of Technetium-99 is approximately 211000 years. The decay of isotopes is represented by the following ordinary differential equation:

[tex]\frac{dm}{dt} = -\frac{m}{\tau}[/tex] (Eq. 1)

Where:

[tex]\frac{dm}{dt}[/tex] - First derivative of isotope mass in time, measured in miligrams per year.

[tex]m[/tex] - Mass of the isotope, measured in miligrams.

[tex]\tau[/tex] - Time constant, measured in years.

Now we proceed to obtain the solution of this differential equation:

[tex]\frac{dm}{m} = -\frac{dt}{\tau}[/tex]

[tex]\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt[/tex]

[tex]\ln m = -\frac{t}{\tau}+C[/tex]

[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)

Where:

[tex]m_{o}[/tex] - Initial mass of the isotope, measured in miligrams.

[tex]t[/tex] - Time, measured in years.

The time passed for isotope is cleared within the equation described above:

[tex]\ln \frac{m}{m_{o}} = -\frac{t}{\tau}[/tex]

[tex]t = -\tau \cdot \ln \frac{m}{m_{o}}[/tex]

In addition, we can obtain the time constant as a function of half-life:

[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex] (Eq. 3)

If we know that [tex]t_{1/2} = 211000\,yr[/tex], [tex]m_{o} = 150\,mg[/tex] and [tex]m = 30\,mg[/tex], then the time passed is:

[tex]\tau = \frac{211000\,yr}{\ln 2}[/tex]

[tex]\tau \approx 304408.654\,yr[/tex]

[tex]t = -(304408.654\,yr)\cdot \ln \left(\frac{30\,mg}{150\,mg} \right)[/tex]

[tex]t \approx 489926.829\,yr[/tex]

The amount of passed half-lives is that time divided by a half-life. That is:

[tex]n = \frac{489926.829\,yr}{211000\,yr}[/tex]

[tex]n = 2.322[/tex]

2.322 half-lives have passed to decay down from 150 miligrams to 30 miligrams.

Explanation:

subtract it into 273.15to convert it in to celcius

Answer:

You forgot to include the diagram use snipping tool to take a picture and upload it

Explanation:

Answer:

Ca2+(aq)   +  2OH-(aq)  →  Ca(OH) 2 ( s )

Explanation:

The balanced equation is given as;

Ca(NO 3 ) 2 ( aq )  +  2KOH(aq)  →  Ca(OH) 2 ( s )  +  2KNO 3 ( aq )

The net ionic equation features only ions that actually participated in the reaction. These ions are identified due to their change in oxidation numbers. Also, the aqueous compounds are broken down into ions.

Ca2+  + 2NO3-  +  2K+  +  2OH-  →  Ca(OH) 2 ( s ) +  2K+   +  2NO3-

Ions present in both sides of the equation above. These are; NO3-,  K+ and OH-

The net ionic equation is;

Ca2+(aq)   +  2OH-(aq)  →  Ca(OH) 2 ( s )

Answer:

The specific heat of the mineral is 0.1272J/g°C

Explanation:

The sample is given energy to the calorimeter and the sample of water.

The energy released for the sample is equal to the energy absorbed for both the calorimeter and the water:

C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT

Where C is specific heat

m is mass of the sample and water

And ΔT is change in temperature

C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)

C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

The specific heat of the mineral is 0.1272J/g°C

The energy released for the sample should be equivalent to the energy absorbed for both the calorimeter and the water:

So,

C(Sample)*m*ΔT = C(Calorimeter)*ΔT + C(water)*m*ΔT

here C is specific heat

m is mass of the sample and water

And ΔT is change in temperature

Now

C(Sample)*149g*(92.7°C-23.7°C) = 12.8J/K*(23.7°C-20.0°C) + 4.184J/g°C*81.4g*(23.7°C-20.0°C)

C(Sample)*10281g°C = 47.36J + 1260.1J

C(Sample) = 0.1272J/g°C

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