what is the gravitational force acting on a 59kg person due to another 59kg person standing 2 m away? (G=6.67x10^-11 m^2/kg^2)

17.99 g of water will formed when nitric acid (63 g) and sodium hydroxide (60 g) are mixed.

When nitric acid and sodium hydroxide are mixed, a neutralisation reaction takes place, with the salt created by the two substances and water serving as the reaction result. The rational response to the statement in this instance is:

Na(OH) + HNO3 ⇒ NaNO3 + H2O

• The formula for HNO3's molar mass is 3xmO + mH + mN, or 3x15.99g + 1g + 14g = 62.97 g/mol.

• The formula for NaOH molar mass is: mO + mH + mNa = 15.99g + 1.00g + 22.99g = 39.98 g / mol.

18g of water and 63g of HNO3

17.99 g of water from 62.97 g of HNO3

62.97 g divided by 18 g and 63 g equals 17.99 g of water.

As a result, 17.99 g of water will form.

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An n-input nmos nor gate has ks = 4mA/V2, kl = 2 mA/V2, vt = 1.0V, VDD = 5.0V. Find the approximate values for VOH and VOL for n = 1, 2 and 3 inputs. QL = sat and QS = ohmic, VI = VOH. For a NOR gate, when all inputs are high, the output is low.

When any input is low, the output is high. Here, it is given that QL is in saturation and QS is in the ohmic region. The relation between VDS and VGS for saturation and ohmic region is given as;$$V_{{DS}} \geq V_{{GS}} - V_{{th}}$$ $$V_{{DS}} \lt V_{{GS}} - V_{{th}}$$where, Vth is the threshold voltage. Also, in saturation region,$$I_{{D}} = \frac{1}{2} K_{{n}} \frac{W}{L} (V_{{GS}} - V_{{th}})^2 $$where, ID is the drain current, Kn is the process parameter (µnCox), W is the width, L is the length of the MOSFET. The value of VOH can be calculated for n = 1 input as follows:To obtain VOH, we need to make all inputs high. Therefore,$$I_{{D}} = \frac{1}{2} K_{{n}} \frac{W}{L} (V_{{DD}} - V_{{th}})^2 $$Substituting the given values, we get,$$I_{{D}} = \frac{1}{2} \cdot 4 \cdot 10^{-3} \cdot \frac{1}{2} (5 - 1)^2 = 16 \mu A $$.

When QL is in saturation region,$$V_{{D}} = V_{{DD}} - I_{{D}}R_{{D}} = 5 - 16 \cdot 10^{-6} \cdot 1.5 \cdot 10^{3} = 2.76V $$Since all the inputs are high and the output is low, VOH = 0.The value of VOL can be calculated as follows:Let us consider n = 2 inputs. In this case, for the MOSFETs in the saturation region,$$I_{{D}} = \frac{1}{2} K_{{n}} \frac{W}{L} (V_{{GS}} - V_{{th}})^2 $$Therefore,$$I_{{D}} = \frac{1}{2} \cdot 4 \cdot 10^{-3} \cdot \frac{1}{2} (5 - 1)^2 = 16 \mu A $$and $$V_{{GS}} = V_{{I}} = V_{{OH}} $$Assuming the MOSFET in the ohmic region is in cutoff state,$$V_{{D}} = V_{{I}} = V_{{OH}} $$Therefore, the output voltage is the voltage drop across the resistor and the MOSFET in the saturation region.$$V_{{OL}} = V_{{D}} + I_{{D}}R_{{D}} = 5 - 16 \cdot 10^{-6} \cdot 1.5 \cdot 10^{3} = 2.76V $$The value of VOH and VOL can be calculated for n = 3 inputs in a similar way.

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If the hydrogen gas is combusted and produces 645L of water vapor at 400 kelvin and 1.75 atm, 2796.96 g mass of the zinc is produced .

Using the ideal gas law equation:

PV = nRT

n = (PV) / (RT)

= (1.75 atm * 645 L) / (0.0821 atm·L/(mol·K) * 400 K)

= 42.71 moles

the balanced equation for the reaction between zinc and hydrochloric acid:

Zn + 2HCl -> \(ZnCl_{2}\)  + \(H_{2}\)

1 mole of zinc produces 1 mole of hydrogen gas. Therefore, the moles of zinc are also 42.71.

The molar mass of zinc is 65.38 g/mol.

Mass of zinc = moles of zinc * molar mass of zinc

= 42.71 moles * 65.38 g/mol

= 2796.96 g

Therefore, the mass of the zinc is 2796.96 grams.

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The equilibrium chemical equation Rb(s) + Cl₂(g) RbCl₂(s) has the appropriate product(s) for the reactivity of rubidium.

Magnesium chloride (MgCl₂) is created when the elements magnesium (Mg) and chlorine (Cl₂) mix. As a result, the reaction is a combination reaction, and the equation that balances it is as follows: Mg(s)+Cl₂(g)→MgCl₂(s)

For instance, salt chlorate, a typical household bleach, is created when chlorine, Cl₂, interacts with sodium hydroxide, NaOH, in a disproportionation process. Sodium and chlorine combine to make sodium chloride, a common table salt.

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The energy of the 488 nm photons emitted by an argon-ion laser is d. 4.07 × 10 − 19 J.

The energy (E) of a photon can be calculated using the equation E = hc/λ, where h is Planck's constant (6.626 × 10^-34 J s), c is the speed of light (2.998 × 10^8 m/s), and λ is the wavelength of the photon in meters.
Converting the wavelength of the photon from nanometers to meters (488 nm = 4.88 × 10^-7 m) and plugging it into the equation, we get:
E = (6.626 × 10^-34 J s)(2.998 × 10^8 m/s)/(4.88 × 10^-7 m)
E = 4.07 × 10^-19 J
Thus, the energy of the 488 nm photons emitted by an argon-ion laser is d. 4.07 × 10 − 19 J.

The energy of a photon can be calculated using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. By converting the wavelength of the photon from nanometers to meters and plugging it into the equation, we can calculate the energy of the photon in Joules.

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The word Solubility means the number of grams of the solute dissolved in 100g of solvent to prepare a saturated solution at a particular temperature.

Here, are few general principles of Solubility mentioned below :

⇒ Polar substances are soluble in polar solvents. Ionic solids and polar covalent compounds are soluble in water, e.g,  KCL, \(Na_{2}CO_{3}\) , Sugar and alcohol are soluble in water.

⇒ Non Polar substances are soluble in water. Non Polar covalent compounds are not soluble in water such as ether, benzene and petrol are insoluble in water.  

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If acetone has a density of 0.7857 \(\frac{g}{cm^{3} }\)  the mass in grams of point A is 22.4 g and the volume at point B is 8.32 mL.

Acetone is known as a chemical substance that is usually found in the environment but can also be produced artificially. Acetone is a polar organic product that interacts very well with water molecules, generating dipole-dipole relationships.It is colorless with a distinctive smell and taste, we find it in products known as cleaning and personal care products, but we can also use it as a solvent for substances.

Also in the environment in plants, trees and in volcano emissions or in forest fires, it does not become toxic in low doses but if it is exposed to an individual in high doses it can become fatal.

In the statement we can find that acetone has a density of 0.7857 \(\frac{g}{cm^{3} }\).

Therefore, we can confirm that if acetone has a density of 0.7857 \(\frac{g}{cm^{3} }\)  the mass in grams of point A is 22.4 g and the volume at point B is 8.32 mL.

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There are 4 moles of spectator ions that remain in solution.

The equation of the reaction is;

Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)

We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.

For Na2CO3

1 mole of Na2CO3 yields 2 moles of NaNO3

3 moles of Na2CO3 yields  3 × 2/1 = 6 moles of NaNO3

For Pb(NO3)2

1 mole of Pb(NO3)2 yields 2 moles of NaNO3

2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3

We can see that Pb(NO3)2 is the limiting reactant.

Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.

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The correct answer is a) Henry's Law. This law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas in the overlying space.

This means that as the partial pressure of the gas increases, more gas molecules will dissolve in the liquid. Henry's Law is important in many areas of science, including chemistry, environmental science, and biology.

For example, it is used to understand the behavior of gases in the atmosphere and their impact on climate change, as well as the ability of aquatic organisms to obtain oxygen from water.

Henry's Law can also be applied to industrial processes such as gas purification and carbonation of beverages.

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\(\begin{aligned} -5x+4y &= 3\\\\ x&=2y-15 \end{aligned}\)

Answer:

Explanation:

Percent composition tells you by mass what percent of each element is present in a compound. A chemical compound is the combination of two or more elements. ... The equation for percent composition is (mass of element/molecular mass) x 100.

Hope it helps:)

This issue can be resolved using the coupled gas law:

P1V1 T1 P2V2 T2where P1, V1, and T1 stand for the beginning pressure, volume, and temperature while P2, V2, and T2 stand for the ultimate pressure, volume, and temperature.When the given values are added to the equation, we obtain:2.23 atm (16.8 L)/T1 = 5.26 atm (11.6 L)/313 KWhen we simplify the equation and find T1, we obtain:

T1 is equal to (2.23 atm x (16.8 L x 313 K) / (5.26 atm x 11.6 L).

= 404.7 KThe gas's starting temperature was therefore roughly 404.7 K
the about some tere temperture while about addedfe therd  tere and some one where so that value time stand

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The Neutral Gator program used methods that reduced the amount of carbon dioxide already in the atmosphere and also helps in decreased production of carbon dioxide.

The rate of CO2 increase is in the last few hundred years is 10 times more with the rate of increase (or decrease, for that matter) in the preceding 400,000 years.

There are many possible reasons for this cause , some primary factors are listed below:

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The mass percent composition of nitrogen in NH3 (Ammonia) is determined by dividing the mass of nitrogen present in one molecule of NH3 by the molar mass of NH3.

Determine the molar mass of NH3. Molar mass of NH3 = (1 × atomic mass of N) + (3 × atomic mass of H)= (1 × 14.01 g/mol) + (3 × 1.01 g/mol)= 17.04 g/mol. Determine the mass of nitrogen in NH3. Since NH3 has one nitrogen atom, the mass of nitrogen present in one molecule of NH3 is 14.01 g/mol.

Calculate the mass percent composition of nitrogen in NH3. Mass percent composition of nitrogen in NH3 = (mass of nitrogen present in NH3 / molar mass of NH3) × 100%= (14.01 g/mol / 17.04 g/mol) × 100%= 82.3 %. The mass percent composition of nitrogen in NH3 is 82.3 %.

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The chemical has a percent composition of 56.35% oxygen and 43.65% phosphorus.

A substance has an oxygen content of 88.79% and a hydrogen content of 11.19%. Calculate this compound's empirical formula. Nonetheless, it's customary to employ the atom ratio with the smallest whole number. H 2 O is the compound's empirical formula as a result.

P mass of 30.98 g and O mass of 40.00 g add up to 70.98 g.

Phosphorus: \((30.98 g / 70.98 g) x 100% = 43.65%\)

\((40 g/70.98 g) x 100% = 56.35% for oxygen.\)

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True. Mutations in mitochondrial genes, specifically in the genes MT-ND1, MT-ND4, MT-ND4L, and MT-ND6, have been linked to Leber's hereditary optic neuropathy (LHON).

Leber's hereditary optic neuropathy (LHON) is a rare genetic disorder that causes vision loss. It is caused by mutations in mitochondrial genes, specifically in the genes MT-ND1, MT-ND4, MT-ND4L, and MT-ND6. These genes provide instructions for making proteins involved in the mitochondria's electron transport chain, which generates energy for the cell. The mutations disrupt the normal function of the electron transport chain, leading to oxidative stress and damage to the cells in the optic nerve. LHON is typically inherited through the mother and can lead to irreversible vision loss.

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The freezing point of a solution depends on the concentration of the solute in the solution. To determine the freezing point of a solution prepared by dissolving 6.423 g of ethanol, ch3ch2oh (molecular weight of 46.07 g/mol), we need to know the mass of the solvent and the freezing point depression constant for the solvent.

Assuming that the solvent is water, which has a freezing point depression constant of 1.86 °C/m, and that the mass of the solvent is 100 g, we can calculate the molality of the solution to be 6.423 g/46.07 g/mol = 0.1393 mol. Using the freezing point depression formula, ΔTf = Kf × m, where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution, we can calculate the freezing point depression to be ΔTf = 1.86 °C/m × 0.1393 mol/kg = 0.259 °C. Therefore, the freezing point of the solution is the freezing point of water (0 °C) minus the freezing point depression (0.259 °C), which is -0.259 °C.

The freezing point of a solution prepared by dissolving 6.423 g of ethanol (CH3CH2OH) with a molecular weight of 46.07 g/mol depends on the solvent used. Ethanol is known to lower the freezing point of the solution due to its effect as a solute. To determine the exact freezing point, one needs to know the solvent, its freezing point, and the molality of the solution. Using the colligative properties formula, ΔTf = Kf * molality, and the freezing point depression constant (Kf) of the solvent, the freezing point depression (ΔTf) can be calculated. Add this to the solvent's freezing point to get the solution's freezing point.

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Answer: This satisfies the law of definite proportion or constant proportion and it can be proven below.

Explanation:

The law of constant or definite proportion states that all pure samples of the same compound no matter how they are made; are made up of the same elements combined in the same proportion by mass. The can be verified by finding the ratio of copper to oxygen in the both experiment listed above.

In the first experiment:

1.37 g of cupric oxide contains 1.098 g copper therefore oxygen = 1.37 − 1.098

=0.272 g

The ratio of copper to oxygen is=

1.098 ÷ 0.272

= 4:1

In the SECOND or next experiment:

1.476 g of cupric oxide contains 1.17 g copper Therefore oxygen = 1.476−1.17

=0.306g

The ratio of copper to oxygen is =

=1.176 ÷ 0.306

=3.8:1

Which is approximately≃4:1.

In conclusion, despite the different methods of the experiments, composition of copper of copper in the two different samples are the same and the cupric oxide obtained from different sources contain the same elements ( copper and oxygen) combined in the SAME ratio.

Answer:

Explanations:

Given the operation below:

Grouping the operation into standard and exponential values will give:

Simplify the result

Write the resulting product in standard form:

Hence the coefficient (green) is 4.3615 and the exponent (yellow) is -1.

Answer:

175 meters

Explanation:

As it's velocity is 35m/s

so,

by formula

35 *5

175 meters

Answer:

Explanation: I did the work online because I just worked on this.

Answer:

I kinda forgot. I'm sorry if I didn't answer your question.

Explanation:

Answer:

A

Explanation:

Answer:

A. Hard

Explanation: got it correct on Edg

Answer:

granulated sugar

Explanation:

the rate of dissolving can be increased by mixing stirring and agitating the

solution. hope this answers your question

Aat 25 °C, the number of dissociated H+ ions there are in 325 ml of an aqueous solution whose pH is 11.41 is 1.235 × 10⁻¹².

The pH scale is a logarithmic measure of the acidity or basicity of an aqueous solution. The pH scale ranges from 0 to 14, with 7 being neutral, below 7 being acidic, and above 7 being basic or alkaline. Therefore, we can use the relationship between pH and [H+] concentration to calculate the number of dissociated H+ ions in an aqueous solution. The relationship is as follows:

pH = -log[H+]

or [H+] = 10^-pH

We can use this relationship to find the number of dissociated H+ ions in 325 mL of an aqueous solution whose pH is 11.41 as follows:

First, we need to find the [H+] concentration from the given pH:

[H+] = 10^-pH = 10^-11.41 = 3.80 × 10⁻¹² M

Now that we know the [H+] concentration, we can calculate the number of dissociated H+ ions using the following formula:

Number of dissociated H+ ions = [H+] × volume of solution in Liters

The volume of the solution is given in milliliters (mL), so we need to convert it to liters (L):

325 mL = 325/1000 L = 0.325 L

Now we can use the formula above to find the number of dissociated H+ ions:

Number of dissociated H+ ions = [H+] × volume of solution in Liters = 3.80 × 10⁻¹² M × 0.325 L = 1.235 × 10⁻¹² mol

The answer is: 1.235 × 10⁻¹² dissociated H+ ions.

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The radius of pd atom will be 137.158 pm

Density of unit cell is given as 12.0 g/cm3

Molar Mass is 106.4 g/mol

We know that

The radius will be calculated by density formula since Pd crystallises in FCC structure , the number of cell per unit cell is 4 and the molar mass is given as 106.4g/mol so the  edge length will be calculated as follows

Density= Z× M/ Na × a3

where,

Z= No of atoms per unt cell

M= Molar mass

Na=  Avogadro's number

a= edge length

12.0= 4× 106.4/ 6.023× 10²³ × a3

a³= 425.6/ 6.023× 10²³ × 12

a³= 5.88×10-²³

a= 3.88× 10^-8

r= √2a/4

r= √2 × 388pm/4

r=137.158pm

r=0.137 nm

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The pd atom will have a radius of 137.158 pm. It is provided that Pd crystallizes in an FCC structure, the radius will be computed using the density formula, there are 4 cells per unit cell, and the molar mass is given as 106.4g/mol, the edge length will be determined as follows:

The unit cell's density is 12.0 g/cm3.

106.4 g/mol is the molar mass.

We know that ,

density = ZM/ Na× a³

where,

Z=Atoms in each unit cell

Molar mass

Avogadro's number is Na.

a = edge length

12.0= 4× 106.4/ 6.023× 10²³ × a³

a³= 425.6/ 6.023× 10²³ × 12

a³= 5.88×10⁻²³

a= 3.88× 10⁻⁸

r= √2a/4

r= √2 × 388pm/4

r=137.158pm

r=0.137 nm

So, the radius of the pd atom will be 0.137nm

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Answer:

113.17 g

Explanation:

This can be solved using the Ideal Gas Equation:

PV=nRT

First step is to calculate the volume of the cylinder in L:

V=πr²h

V= π (11.5)² (140)

V= π (132.25) (140)

V = π (18.515)

V = 58,166.588 cm³ = 58.2 L

Then, convert kpa into atm:

1 kpa = 0.00987 atm

108 kpa = 1.06588 atm

Then, plug all of your values into the Ideal Gas Equation to solve for moles:

(1.06588)(58.2)=n(0.082057)(294)

62.034=n(24.125)

2.571357513=n

Finally, convert the moles into grams

2N = 28.0134 amu

O = 15.999 amu

_______________

44.0124 amu

(2.571357513moles)(44.0124 amu) = 113.17 g