Balance the N-14(d,n) nuclear equation for radioisotope production in PET. This means to
write it out in equation form with 2 reactants and 2 products.

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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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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Answer:

Explanation:

w = 16.0KJ

your welcome for the answer let me know ifyou need anymore help.

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Answer:

In pair NaF and H2O both compounds exibit predominantly ionic bonding.

NaF and H[tex]_2[/tex]O exhibit predominantly ionic bonding. Therefore, the correct option is option D among all the given options.

Ionic bonding produces ionic, or electrovalent, combinations, which are best represented by compounds created between nonmetals and alkali as well as alkaline-earth metals.

The electrostatic attraction forces among opposing charges as well as repulsion between comparable charges position the ions in such a way that every positive charge is surround by negative ions as well as vice versa within ionic crystal structures of such a type.

In summary, the ions are organized in such a way that negative and positive charges alternate constantly balance one another, resulting in a total charge of zero for the entire material. NaF and H[tex]_2[/tex]O exhibit predominantly ionic bonding.

Therefore, the correct option is option D.

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Answer:

this is an Oxidation-reduction

Explanation:

like is w reducing agent, N2 is an oxidizing agent

The final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L.

The volume of a given gas can be calculated using the Charles law equation as follows:

V1/T1 = V2/T2

Where;

2/248 = V2/273

273 × 2 = 248V2

546 = 248V2

V2 = 546/248

V2 = 2.2L

Therefore, the final volume of the gas that was heated from -25.0 °C to standard temperature is 2.2L

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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< !!!  

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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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

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

Answer:

hydrocarbon

Explanation:

it reacts with oxygen to produce water and carbon dioxide.

hope this helps.

Answer:

molecular mass of each substance

molecular mass of each substance

Explanation:

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:

H30,+ion is known as Hydroniom Ion

Explanation:

brain ly ----

everyday

brainless this answer is correct

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:

I hope it helps mark me brainliest^^

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.

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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The following sets of quantum numbers ( n, I, ml, ms ) that cannot occur within an atom of gallium are :

Given that

Ga ( 31 ) = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p¹

                 [ 4r ] 3d¹⁰ 4s²4p¹

where : n = principal quantum numbers,  l = azimuthal quantum numbers, ml = magnetic quantum numbers, ms = spin quantum numbers.

For option B

Given that 4d shell is absent for Gallium the set cannot occur within the atom of Gallium

For Option D and E

All quantum number cannot occur here because the value of l = 2 for option D while for option E the value of l = 3

Hence we can conclude that sets of quantum numbers ( n, I, ml, ms ) that cannot occur within an atom of gallium are as listed above.

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why did u gave answer before

Explanation:

1) Weather refers to temporary atmospheric conditions that are occurring at a given point and time. Climate refers to the average weather conditions that occur in an area over an extended period.

2) Regions nearer the equator receives sunlight more directly than regions located farther from the equator. As a result, regions near the equator experience warm temperatures all year. Regions located farther from the equator experience more temperature fluctuations.

3) An air mass containing extremely cold, dry air would likely be classified as a continental arctic air mass.

4) The ITCZ is the point at which trade winds from the Northern Hemisphere converge with trade winds from the Southern Hemisphere. This area typically experiences calm weather.

5) Polar jet streams occur at higher latitudes. They separate the polar easterlies from the prevailing westerlies. Tropical jet streams occur at lower latitudes. They separate the trade winds from the prevailing westerlies.

Answer:

Explanation:

Pb(NO3)2(aq) + 2NaCl(aq) ==> PbCl2(s) + 2NaNO3(aq)

NO3- mixes with everything at beginning levels.

PbCl2 doesn't mix at all in water. It becomes a ppte, which means it is solid form. Your teacher might prefer using PbCl2(ppte). I think I'd ask to make sure.

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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Answer:

A. Rubidium and phosphorus react to form rubidium phosphide

Explanation:

P is not potassuim. It is phosporous. So A is the only correct choice.

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]

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:

probably the last one should hbthrbwhdee.

Answer:

it brings oxygen into the blood and removes carbon dioxide

Explanation:

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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Answer:

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

Explanation:

A coarse particle has a diameter of 1x10-6m. a nanoparticle has a diameter of 1.6 x 10-9m. Calculate how many times bigger the diameter of the coarse particle is than the diameter of the nanoparticle

Answer:

h = 384.4 m

t = 8.5 s

Explanation:

The firework will explode at the highest point. Because it would start to move downward at the highest point. Using 3rd equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration due to gravity at that planet = -10.56 m/s²

h = maximum height = ?

Vf = Final velocity = 0 m/s (since, firework stops at highest point)

Vi = Initial Velocity = 90.1 m/s

Therefore,

2(-10.56 m/s²)(h) = (0 m/s)² - (90.1 m/s)²

h = 384.4 m

Now, we use 1st equation of motion:

Vf = Vi + at

0 m/s = 90.1 m/s + (-10.56 m/s²)(t)

t = 8.5 ss

Answer:

∆H or Enthalpy of the reaction

Explanation:

If ∆H is +ve

If ∆H is -ve