what is hydrogen alkali metal or halogen​

Intensive properties do not change as the size of an object changes but Extensive properties change as the size of an object changes; here the properties can not change because the amount of the water stays the same

The physical properties of matter can be divided into  intensive and extensive properties and the terms are introduced in 1917 by Richard C Tolman.

Intensive properties are independent of the presence of number of substance and they are bulk properties, here the Characteristic doesn’t change.

The size of intensive properties cannot change, examples are Density, temperature, pressure etc. It can not depend on the amount of matter and also intensive properties are fixed, countable.

Extensive properties are dependent on the presence of  number of substance, easily identified, Size of Extensive properties changes and  It can be computed.

extensive properties depend on the amount of matter, often changes and are not countable; For example Volume, size, mass, length, weight

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Answer:

im guessing its A run off

Explanation:

Answer: Run off

Explanation:

Wetlands have plants that clean water through. As water flows into a wetland it encounters the the plants growing there. This slows the water down making it less likely to cause erosion. The nutrient pollutants nitrogen and phosphorus are absorbed by the roots of the plants.

Answer:

the correct answer is H2SO4 and HSO4-41

Answer:

When both liquid molecules are polar then they can attract one another - which leads to mixing (miscibility).

When the molecular liquid is nonpolar, then the water molecules attract only one another while ignoring the nonpolar liquid. the result is that the two liquids are immiscible.

Explanation:

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Answer:

They lack lack inter-molecular forces to disolve in the polar liquids.

Explanation:

Miscibility is the property of two substances to mix in all proportions (that is, to fully dissolve in each other at any concentration). Polar molecules like water will mix well with some other polar molecules but things like oil which are non polar do not mix well with polar molecules. This is because the oil and water lack inter-molecular forces to keep them mixed.

Answer:

[tex]\displaystyle K_a = 4.24\times 10^{-10}[/tex]

Explanation:

Write the base reaction of NaA with water:

[tex]\displaystyle \text{A}^-_\text{(aq)}+\text{H$_2$O}_\text{($\ell$)}\rightleftharpoons \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)}[/tex]

Hence, the equilibrium constant expression for the reaction is:
[tex]\displaystyle K_b = \frac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]}[/tex]

Thus, to find Ka, we can find Kb and use the fact that Ka × Kb = Kw.

From the reaction and initial concentration of NaA, create an ICE chart:

[tex]\begin{tabular}{llllll} & A^- &\text{H$_2$O} & \rightleftharpoons & HA & OH^- \\I & 0.563 M & \---- & & 0 M & 0 M \\C & -\text{ $ x$} & \---- & & +\text{ $x$ M} & + \text{$x$ M} \\E & \text{(0.563 - $x$) M} & \---- & & \text{$x$ M} & \text{$x$ M} \end{tabular}[/tex]

Find [OH⁻] from the given pH:
[tex]\displaystyle \begin{aligned} \text{pH} +\text{pOH} & = 14.00 \\ \\ \text{pOH} & = 14.00 - \text{pH} \\ \\ & = 14.00 - (11.56) \\ \\ & = 2.44 \\ \\ -\log[\text{OH}^-] & = 2.44 \\ \\ [\text{OH}^-] &= 10^{-2.44} \\ \\ & =0.00363 \text{ M}= 3.63\times 10^{-3} \text{ M} = x\text{ M}\end{aligned}[/tex]

Solve for all species concentrations at equilibrium from the found x value:

[tex]\displaystyle [\text{HA}] = [\text{OH}^-] = 3.63\times 10^{-3} \text{ M}[/tex]

And:

[tex]\displaystyle \begin{aligned} \ [\text{A}^-] & = 0.563 - 3.63\times 10^{-3} \text{ M}\\ \\ & = 0.559\text{ M}\end{aligned}[/tex]

Find Kb:

[tex]\displaystyle \begin{aligned} \displaystyle K_b &= \frac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]} \\ \\ & = \frac{(3.63\times 10^{-3})(3.63\times 10^{-3})}{(0.559)}\\ \\ & = 2.36\times 10^{-5}\end{aligned}[/tex]

Find Ka:

[tex]\displaystyle \begin{aligned} K_a\cdot K_b & = K_w \\ \\ K_a & = \frac{K_w}{K_b} \\ \\ & = \frac{(1.00 \times 10^{-14})}{(2.36\times 10^{-5})} \\ \\ &= 4.24\times 10^{-10} \end{aligned}[/tex]

In conclusion:
[tex]\displaystyle K_a = 4.24\times 10^{-10}[/tex]

[tex]pH = - \log[\text{H}^{+}] = -\log \left(1.75 \times 10^{-11} \right) = 10.757[/tex]

The mass of the salt only if 35.9564g is the combined mass of the beaker and the salt is 8.7224g

According to this question, an empty beaker has a measured mass of 27.234g. However, when some salt is added to the beaker, the combined mass is 35.9564g.

This means that to calculate the mass of the salt in the beaker, we have to subtract the mass of the empty beaker from the combined mass as follows:

mass of salt only = 35.9564g - 27.234g

mass of salt only = 8.7224g

Therefore, the mass of the salt only if 35.9564g is the combined mass of the beaker and the salt is 8.7224g.

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Answer:

Explanation:

w = 16.0KJ

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If he rubs his foot on the carpet or touches the metal doorknob it would increase his chances of being zapped.

Rubbing his feet would be friction.

Touching the doorknob would be potential energy.

Side note: I love the format for this question, was not expecting to see John Travolatge on here today >O< !!!  

Answer:

this is an Oxidation-reduction

Explanation:

like is w reducing agent, N2 is an oxidizing agent

For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K  is mathematically given as

Ecell = 0.577 V

Generally, the equation for the Oxidation and Reduction  is mathematically given as

Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O

Sn+2(aq) + 2e- ------------ Sn(s) ] x 3  ...R

Reaction

 2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)

Therefore

Eicell = - 0.14 - ( - 0.74)

Eicell = 0.60

In conclusion

[tex]Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}[/tex]

[tex]Ecell = 0.60 - \frac{0.0591 }{6} \frac{log( 1.20)^2}{ ( 0.200)^3}[/tex]

Ecell = 0.577 V

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Answer:

H30,+ion is known as Hydroniom Ion

Explanation:

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brainless this answer is correct

Answer:

Na+

Explanation:

it is least pull by the nuclear charge

Answer:

Na+

Explanation:

It is posetively charged and in the first column

The volume in litres occupied by 1.59 Kg of ethyl alcohol given the data from the question is 2.03 L

Mole = mass / molar mass

Mole of ethyl alcohol = 78.5 / 46.07

Mole of ethyl alcohol = 1.7 mole

Mole = mass / molar mass

Mole of ethyl alcohol = 1590 / 46.07

Mole of ethyl alcohol = 34.5 moles

V₁ / n₁ = V₂ / n₂

0.1 / 1.7 = V₂ / 34.5

Cross multiply

1.7 × V₂ = 0.1 × 34.5

Divide both side by 1.7

V₂ = (0.1 × 34.5) / 1.7

V₂ = 2.03 L

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Explanation:

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Atomic weight of nitrogen is 23

Answer:

see below

Explanation:

.3 M/L ( .125 L) = .0375 mole required

.0375 mole / 2 mole / l = .01875 l required

take 18.75 ml of the 2.0 solution then dilute it to 125 ml

For a (unbalanced) reaction: NaOH +CO2-Na2CO3 + H2O, the moles of NaOH and moles of each product are formed are mathematically given as

a) Moles of NaOH =44.05

b)   Moles of Na2CO3=21.0

Generally, the equation for the Chemical reaction  is mathematically given as

2 NaOH(aq)+ CO2(g)------> Na2CO3(aq)+ H2O(l)

Therefore

Moles of CO2= 925/44

Moles of CO2=21.0

Hence

Moles of NaOH = 2 x Moles of CO2

Moles of NaOH = 2x925/44

Moles of NaOH =44.05

In conclusion

Moles of Na2CO3 925/44

Moles of Na2CO3=21.0

And

Moles of H2O= 925/44

Moles of H2O= 21.0

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The density of the rock with a mass of 13 g and is dropped into a graduated cylinder containing 12 mL of water is 0.57g/mL.

The density of a substance can be calculated using the following formula:

Density = mass ÷ volume

According to this question, a small rock with a mass of 13 g is dropped into a graduated cylinder containing 12 mL of water. The water rises to 35 mL.

This means that the volume of the rock is 35 - 12 = 23mL

Density of the rock = 13g ÷ 23mL

Density = 0.57g/mL

Therefore, the density of the rock with a mass of 13 g and is dropped into a graduated cylinder containing 12 mL of water is 0.57g/mL.

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Answer:

Your body can cool itself by sweating. When sweat evaporates, it lowers your temperature

Explanation:

Explanation:

N2 (g) + H2 (g) gives out NH3 (g)

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.

Now balance it. You have two reactants with compositions involving a single element, which makes it very easy to keep track of how much is on each side. I would balance the nitrogens, and then the hydrogens.(If you balance the hydrogen reactant with a whole number first, I can guarantee you that you will have to give NH3 a new stoichiometric coefficient.)

N2 (g) + 3H2 (g) gives out 2NH3 (g)

The stoichiometric coefficients tell you that if we can somehow treat every component in the reaction as the same (like on a per-mol basis, hinthint), then one "[molar] equivalent" of nitrogen yields two [molar] equivalents of ammonia.

Luckily, one mol of anything is equal in quantity to one mol of anything else because the comparison is made in the units of mols.

So what do we do? Convert to

mols (remember the hint?).

28g N2 × 1 mol N2/ 2 × 14.007) g N2

= 0.9995 mol N2

At this point you don't even need to calculate the number of mols of H2 . Why? Because H2 is about 2 g/mol, which means we have over 10 mols of H2. We have 1 mol N2, and we need three times as many mols of H2 as we have

N2.

After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2.

Thus, we should yield 2×0.9995=1.9990 mols of NH3 (refer back to the reaction). So this is the second and last calculation we need to do:

1.9990 mol NH3 × 17.0307 g NH3/ 1 mol NH3

= 34.0444 g NH3

Hope it helpz~

34grams of NH₃ can be produced from the reaction of 28g of N₂ and 25g of H₂ and 19g of the excess reagent is left over.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Mass of N₂ = 28g

Mass of H₂ = 25g

Moles of N₂ = mass / molar mass

                    = 28 / 28

                   = 1 mole

Moles of H₂ = 25 / 2

                     = 12.5 moles

Since the number of moles of N₂ is lesser, the amount of products formed depends on the moles of N₂.

From the reaction,

1 mole of N₂ gives 2 moles of NH₃

So, Mass of NH₃ in 2 moles = 2 × 17

                                              = 34g

B) From the reaction,

1 mole of N₂ reacts with 3 moles of H₂.

The number of Moles of H₂ left = 12.5 - 3

                                                    = 9.5 moles

Mass of excess reagent left = moles left × molar mass of H₂

                                             = 9.5 × 2

                                             = 19g

Therefore, 34grams of NH₃ can be produced from the reaction of 28g of N₂ and 25g of H₂ and 19g of the excess reagent is left over.

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Take a example of water

Attached structure

Answer:

option B = 0.410 g

Explanation:

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The colors and the corresponding wavelengths the three types of cones detect are:

Human vision refers to the ability of humans to see and visualize objects in the environment in terms in texture, color, pattern, etc.

The cells of the human eye responsible for vision are the rods and cones.

The cones are responsible for color-vision while rods are responsible for black and white vision.

The three types of cones detect blue, green, and red colors corresponding to wavelengths given below:

Therefore, the cones in the eye detect blue, green and red colors.

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The time elapsed when the ball placed above the concave mirror and the image formed would be at the same location is 0.55 s.

The position of the image formed is determined using the followimg mirror formula;

[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex]

where;

f = R/2

f = 1.5/2

f = 0.75 m

When the ball and its image is in the same position, u = v

The position of the ball is calculated as;

[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{u} \\\\\frac{1}{f} = \frac{2}{u} \\\\u = 2f\\\\u = 2(0.75)\\\\u = 1.5 \ m[/tex]

The time taken for the ball to travel the caluclated distance is determined as;

h = ut + ¹/₂gt²

1.5 = 0  + ¹/₂(9.8)t²

1.5 = 4.9t²

t² = 1.5/4.9

t² = 0.306

t = 0.55 s

Thus, the time elapsed when the ball and its image are at the same location is 0.55 s.

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Answer:

d) I, III, and IV

Explanation:

rust, copper oxidation, and percipitate from a chemical reaction are all, well, chemical reactions.

Answer:

∆H or Enthalpy of the reaction

Explanation:

If ∆H is +ve

If ∆H is -ve

Explanation:

density=> ro=m/V

m=0,12kg

V=0,04m³=6,4.10^-5

=>0,12/6,4.10^-5=1875 kg/m³

If 6 units of AB were reacted with 10 units of CD in the described equation, the limiting reactant would AB

They are reactants that determine how far a reaction can go in terms of yield.

From the equation: 3AB + 4CD --> 2AD + 6CB

The mole ratio of AB to CD is 3:4

Thus, 6 units of AB will require 8 units of CD.

But 10 units of CD were reacted with only 6 units of AB. This means that CD is in excess by 2 units while AB will limit the yield of the reaction.

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Answer:

3

Explanation:

PH = -Log [H+]

PH = -Log [1 × 10^-3]

PH = -Log [10^-3]

Log (10) = 1

PH = -(-3) Log (10)

PH = 3 × 1

PH = 3