Is 93 degree Celsius cold or hot

(a) The hazard quotient is:

HQ = Intake / RfD = 0.0002 mg/day / 0.02 mg/kg/day = 0.01

(b) The cancer risk is 1 in 10,000.

(c) The cancer risk is 1 in 1,000.

Hazard quotient (HQ) is a measure used in risk assessment to evaluate the potential health risk posed by exposure to a chemical or other hazard. It is calculated as the ratio of the dose or exposure level of the chemical to a reference dose (RfD) or reference concentration (RfC) established by regulatory agencies or scientific bodies as a safe level of exposure. If the hazard quotient is greater than 1, it suggests that the level of exposure is of potential concern and additional risk assessment may be needed.

(a) The hazard quotient (HQ) is calculated as the daily intake of a chemical divided by its reference dose (RfD). The RfD for 1,1-dichloroethylene is 0.02 mg/kg/day.

The daily intake of 1,1-dichloroethylene can be calculated as:

Intake = concentration × ingestion rate × body weight

Intake = 0.1 μg/L × 2 L/day × 70 kg = 14 μg/day = 0.0002 mg/day

The hazard quotient is:

HQ = Intake / RfD = 0.0002 mg/day / 0.02 mg/kg/day = 0.01

(b) The cancer risk from exposure to 1,1-dichloroethylene can be estimated using the unit risk factor (URF) for this chemical, which is 0.5 per mg/kg/day. The cancer risk is calculated as:

Risk = Intake × URF = 0.0002 mg/day × 0.5 per mg/kg/day = 0.0001

Therefore, the cancer risk is 1 in 10,000.

(c) If the individual drinks this water for 30 years, the total exposure would be:

Exposure = Intake × 365 days/year × 30 years = 2.19 mg

The new hazard quotient is:

HQ = Exposure / (RfD × body weight) = 2.19 mg / (0.02 mg/kg/day × 70 kg) = 1.57

The new cancer risk is:

Risk = Exposure × URF = 2.19 mg × 0.5 per mg/kg/day = 1.10

Therefore, the cancer risk is 1 in 1,000.

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Answer:

It's the A. 2-bromobutane

Explanation:

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The name of the alkane molecule \(\rm CH_3CH_2CHBrCH_3\) is 2-bromobutane. The correct answer is option A.

An alkane is a type of hydrocarbon molecule that consists entirely of single-bonded carbon and hydrogen atoms. Alkanes are also known as saturated hydrocarbons.

Therefore, 2-bromobutane is the name of the alkane molecule \(\rm CH_3CH_2CHBrCH_3\). Option A is the correct answer.

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Answer:

Approximately \(0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}\).

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

\(\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C\).

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took \(2911\; \rm J\) of energy to raise the temperature of this sample by \(\Delta T = 92.6\; \rm ^\circ\! C\). Therefore, raising the temperature of this sample by \(1\; \rm ^\circ\! C\) (unit temperature) would take only \(\displaystyle \frac{1}{92.6}\) as much energy. That corresponds to approximately \(31.436\; \rm J\) of energy.

On the other hand, the energy required to raise the temperature of this material by \(1\; \rm ^\circ\! C\) is proportional to the mass of the sample (also assuming no phase change.)

It took approximately \(31.436\; \rm J\) of energy to raise the temperature of \(57.07\; \rm g\) of this material by \(1\; \rm ^\circ C\). Therefore, it would take only \(\displaystyle \frac{1}{57.07}\) as much energy to raise the temperature of \(1\; \rm g\) (unit mass) of this material by \(1\; \rm ^\circ \! C\!\). That corresponds to approximately \(0.551\; \rm J\) of energy.

In other words, it takes approximately \(0.551\; \rm J\) to raise \(1\; \rm g\) (unit mass) of this material by \(1\; \rm ^\circ \! C\). Therefore, by definition, the specific heat of this material would be approximately \(0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}\).

Answer:So, about 0.093 g of the isotope will remain after 19.8 days.

Explanation:

The first step is to find the number of half-lives that have passed during 19.8 days:

Number of half-lives = time elapsed / half-life

Number of half-lives = 19.8 days / 3.8 days per half-life

Number of half-lives ≈ 5.21

This means that the initial amount of the isotope has been halved 5.21 times. The remaining fraction of the original amount can be calculated using the following formula:

Remaining fraction = (1/2)^(number of half-lives)

Substituting the values, we get:

Remaining fraction = (1/2)^5.21

Remaining fraction ≈ 0.031

Therefore, the amount of the isotope remaining after 19.8 days is:

Remaining amount = Remaining fraction x Initial amount

Remaining amount = 0.031 x 3.00 g

Remaining amount ≈ 0.093 g

So, about 0.093 g of the isotope will remain after 19.8 days.

The isotope in grams will remain after 19.8 days would be 0.081 grams.

The formula to calculate the left mass of a radioactive element can be deduced as -

\( \qquad\star\longrightarrow \underline{\boxed{\sf{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½}} }}} \\\)

Where-

According to the given specific parameters -

Now that we have all the required values, so we can plug them into the formula and solve for the left mass of a radioactive element-

\( \qquad \longrightarrow \sf \underline{m =m_{o} \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{t}{T½} }} \\\)

\( \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{19.8}{3.8} } \\\)

\(\qquad \longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ \dfrac{\cancel{19.8}}{\cancel{3.8}} } \\\)

\( \qquad\longrightarrow \sf m =3 \times { \bigg(\dfrac{1}{2} \bigg)}^{ 5.21052..... } \\\)

\( \qquad\longrightarrow \sf m =3 \times 0.02700... \\\)

\( \qquad\longrightarrow \sf m =0.081020....\;g \\\)

\( \qquad\longrightarrow \sf \underline{m =\boxed{\sf{0.081\;g}}} \\\)

Answer: 20.9 billion years.

Explanation: To estimate the age of the rock, we can use the fact that the half-life of uranium-238 is 4.47 billion years. This means that half of the uranium-238 in the rock would have decayed into lead-206 after 4.47 billion years, and half of what remains would decay in the next 4.47 billion years, and so on.

If 20% of the uranium-238 in the rock has decayed into lead-206, then 80% of the original uranium-238 is still present in the rock. This means that the rock has gone through one half-life of uranium-238 decay, and we can estimate its age by using the graph on page 253.

According to the graph, the ratio of lead-206 to uranium-238 after one half-life is approximately 0.027. This means that the rock contains 0.027 times as much uranium-238 as it did originally, and the remaining 80% of the uranium-238 corresponds to 0.8 times the original amount.

Therefore, we can estimate the original amount of uranium-238 in the rock as follows:

Original amount of uranium-238 = (0.8) / 0.027 = 29.63 times the current amount

Next, we can use the fact that each half-life corresponds to a reduction in the amount of uranium-238 by a factor of 2, to estimate the number of half-lives that have passed since the rock formed:

Number of half-lives = log2(29.63) = 4.88

Finally, we can estimate the age of the rock as follows:

Age of the rock = (4.88) x (4.47 billion years per half-life) = 20.9 billion years

Therefore, we can estimate that the age of the rock is approximately 20.9 billion years.

The balanced chemical equation is

Pb(NO3)2 (s) + H2SO4 (aq) PbSO4 (s) + 2HNO3 for the reaction of lead sulphate from lead nitrate solution and dilute

sulfuric acid (aq).

ionic net equation PbSO4 is formed by combining SO4 2-(aq) and Pb 2+(aq) (s)

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The mass of the solid aluminum that forms are 192.73 grams

To determine the mass of solid aluminum that forms, we need to use stoichiometry and the balanced chemical equation for the reaction between magnesium and aluminum nitrate.

The balanced chemical equation is:

3 Mg + 2 Al(\(NO_{3}\))3 → 3 Mg(\(NO_{3}\))2 + 2 Al

The equation shows that 3 moles of magnesium react with 2 moles of aluminum to produce 2 moles of aluminum nitrate.

To calculate the mass of solid aluminum, we need to know the amount of magnesium used. Given that the volume of the magnesium is 150.0 mL and its density is 1.738 g/cm³, we can calculate the mass of magnesium using the formula:

Mass = Volume × Density

Mass of magnesium = 150.0 mL × 1.738 g/cm³ = 260.7 g

Now, using the molar mass of magnesium (24.31 g/mol) and the molar ratio from the balanced equation, we can determine the moles of magnesium used:

Moles of magnesium = Mass of magnesium / Molar mass of magnesium

= 260.7 g / 24.31 g/mol

= 10.72 mol

According to the stoichiometry of the balanced equation, the ratio of moles of magnesium to moles of aluminum is 3:2. Therefore, the moles of aluminum formed will be:

Moles of aluminum = (2/3) × Moles of magnesium

= (2/3) × 10.72 mol

= 7.15 mol

Finally, we can calculate the mass of solid aluminum using its molar mass (26.98 g/mol):

Mass of aluminum = Moles of aluminum × Molar mass of aluminum

= 7.15 mol × 26.98 g/mol

= 192.73 g

Therefore, the mass of the solid aluminum that forms is approximately 192.73 grams.

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0.72 L and 2.28 L of 1/5 M and 1/8 M HCl solution are required respectively to obtain 3 L of 1/7 M HCl.

The concentration of a solution is the amount of substance a given volume of the solution of that substance.

The final concentration and volume of the mixture is 3 L of 1/7 M HCl.

Let the volume of 1/5 M and 1/8 M HCl solution required be X and Y respectively.

X + Y = 3 L

Liters of y = 3 -

[.0.2 X + 0.125 * (3-X)] / 3 = 1/7

(0.2X + 0.375 - 0.125X) / 3 = 1/7

(0.075 X / 3) = 1/7 - 0.375/3

(0.075X / 3) = 0.018

Multiplying both sides by 3

0.075 X = 0.054

X = 0.72 L

Y = 2.28 L

Therefore, 0.72 L and 2.28 L of 1/5 M and 1/8 M HCl solution are required respectively.

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Answer:

proton and neutron in region one and electron orbiting in region two is the answer

Answer:

We need more? What else is in the question? This is unanswerable.

Explanation:

Answer:

The correct answer to the following question will be "4.08 × 10⁻¹⁹ Joule".

Explanation:

Given:

Wavelength, λ = 486.0 nm

As we know,

\(E=h\upsilon =\frac{hc}{\lambda}\)

On putting the estimated values, we get

⇒          \(=\frac{1241.5 \ ev\ nm}{486 \ nm}\)

⇒          \(=2.554 \ ev\)

∴ 1 ev = 1.6 × 10⁻¹⁹ J

Now,

Energy, \(E=2.554\times 1.6\times 10^{-19}\)

⇒               \(=4.08\times 10^{-19} Joule\)

Step 1 - Understanding the relation between temperature and pressure

There are three main variables that can modify the state of a gas: pressure (P), volume (V) and temperature (T). When one of them is kept constant, linear relationships arise between the remaining two.

So when we keep the volume constant (isovolumetric transformation), the pressure and the temperature become directly proportional, i.e., the greater the temperature, the greater the pressure and vice-versa.

We can state it mathematically as:

Step 2 - Using the equation to solve the question

The electron configuration 1s² 2s²2p6 is the element Neon.

One orbital may house a maximum of two electrons, and there are four different types of orbitals (s, p, d, and f). More electrons can be held in the p, d, and f orbitals since they contain various sublevels. As was said, each element's location on the periodic table determines the specific electron configuration of that element.

The octet rule states that an atom's outermost shell may hold no more than 8 electrons (except K shell which can accommodate maximum 2 electrons). Hence, potassium's electrical structure.

The quantity of valance electrons, which take part in the creation of a chemical bond, is found in an atom's outermost shell. The outermost shell of the atom's electrical structure contains 4 valance electrons.

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Answer:

Biodiversity gives resilience—from the microbes that contribute to the formation of the human biome to the genes that help us adapt to stress in the environment

Explanation:

supports all forms of livelihoods, may help regulate disease, and is necessary for physical, mental, and spiritual health and social well-being.

Answer:

Newtonian fluids have a constant viscosity that does not change and non newtonain fluids have a viscosity that varies.

Explanation

Newtonian fluids have a constant viscosity and a zero shear rate at zero shear stress. Non-Newtonian fluids have a variable viscosity and a varible relationship with shear stress.

The volume measurements for the 25ml and 100ml graduated cylinders are :

The average deviation from the mean is calculated thus :

Σ|X - mean| ÷ n

For 25ml cylinder :

Avearage = mean = 19

(|18.5 - 19| + |19.5 - 19| + |19 - 19) ÷ 3 = 0.3333

Volume measurement = 19 ± 0.333

For 100 ml cylinder :

Average = [(19 + 20 + 20) ÷ 3] = 19.67

Avearage = mean = 19.67

(|19 - 19.67| + |20 - 19.67| + |20 - 19.67) ÷ 3 = 0.44

Volume measurement = 19.67 ± 0.443

Therefore, the volume measurement for the two graduated cylinders are : 19 ± 0.333 and 19.67 ± 0.443

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Pyruvate is a byproduct of glycolysis.The oxidative deacetylation of pyruvate to produce carbon dioxide and acetyl CoA is catalyzed by the multienzyme complex dopamine dehydrogenase complex (PDH complex).Pyruvate + CoA + NAD+ ----> Acetate CoA + NADH + H++ CO2. Acetyl CoA ———- Citric Acid Cycle is the general equation for the reaction.

Coenzyme A is joined with the oxidized two-carbon acetyl group to generate acetyl CoA.

The pyruvate dehydrogenase (PDH) enzyme, which is a component of the multienzyme PDC and is frequently described to as a "gatekeeper" in the oxidation of carbohydrates, catalyzes the physiologically irreversible conversion of pyruvate to acetyl-CoA.

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85.4 g/mol s the molar mass of a pure gaseous compound having a density of 4.95 g/L at -35⁰C and 1020 mm Hg.

The mass in grammes of one mole of a chemical is its molar mass. A mole is the measurement of the number of things, such as atoms, molecules, and ions, that are present in a material. Every material has 6.022 x 10²³ molecules in a mole.

ρ = 4.95 g/L = 4.95 kg/m³, the density

T = -35°C = -32+273 K = 238 K, the temperature

The ideal gas law is

p = ρ*(R/M)*T

where

R = 8.314 J/(mol-K), the gas constant

M = molar mass, kg/mol

The molar mass is

M = ρRT / P

M = (4.95 kg/m³) * 8.314 J/(mol-K) * (238 K) / 1.1467 * 10⁵ Pa

M = 0.0854

The molar mass is 0.0854 kg/mol = 85.4 g/mol

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24. The IUPAC name of the compound is 2,4-dimethyl-1-pentanal (Option B)

25. The IUPAC name of the compound is 5-methyl-3-octanone (Option C)

The international union of pure and applied chemistry (IUPAC) standard for naming organic compounds is given below:

24. The name of the compound is given as follow:

Thus, the IUPAC name of the compound is:

2,4-dimethyl-1-pentanal (Option B)

25. The name of the compound is given as follow:

Thus, the IUPAC name of the compound is:

5-methyl-3-octanone (Option C)

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Answer:

1s2 2s2 2p5

is the correct answer

Explanation:

Hi there!

From the figure;

We can figure out that the atomic number is 7.

So;

Option A: 1s² 2s² 2p²

Sum of the powers of variable is just 6. So, it's not answer.

Option B: 1s²2s²2p³

Sum of the powers of variable is 7. So it's the answer

Option C: 1s²2s²2p⁴

Sum of the powers of variable is 8. So, it's not answer.

Option D: 1s²2s²2p⁵

Sum of the powers of variable is 9. So, it's not the correct answer.

Therefore, answer is option B.

Hope it helps!

Answer:

This bubbled-up magma is cooled by frigid seawater to form igneous rock. This rock (basalt) becomes a new part of Earth's crust. Seafloor spreading occurs along mid-ocean ridges—large mountain ranges rising from the ocean floor.

Explanation:

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carry on learning

Answer:

Farmers spray to mitigate crop damage caused by pests. A pest is any biological organism, including weeds, pathogens, and arthropods, that interferes with the production of crops affecting quality and/or yield. ... Pesticides work in many different ways by affecting their target, whether it be a weed, pest, or disease.

Explanation:

this is my answer❤︎

Answer:

I dont get it what words are we using bc I see words on the side just it doesn't make any sense

When temperature of fluid increases, kinetic energy increases, viscosity decreases and the fluid flows faster.

When temperature of a fluid decreases, kinetic energy decreases, viscosity increases and the fluid flows slower.

When temperature of a fluid stays the same, kinetic energy stays the same, viscosity stays the same and the fluid flow stays the same.

Viscosity is the friction that exist between the internal layers of a fluid in motion.

Viscosity decreases with temperature due to the increase in the kinetic energy of the molecules of the fluid.

Based on the relationship between temperature, kinetic energy and viscosity:

Therefore, it can be concluded that viscosity of a fluid depends on temperature.

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B) Planck's constant (h)

Answer:

answer is first one 1 will be less then that of hg coloumn