A sample of gas has an initial volume of 20 L and an initial pressure of 2.5 atm. If the pressure changes to 3.1 atm, what is the new volume, assuming that the temperature remains constant?

The boiling point elevation is a colligative property. From the calculations, the boiling point of the ethanol - sugar solution is 78.65°C.

The term colligative properties refers to those properties that depend on the number of solute particles present.

Number of moles of sugar = 20 g/342.3 g/mol = 0.058 moles

Boiling constant of ethanol = 1.22 °C⋅kg/mol

Molality of the solution = 0.058 moles/250 * 10^-3 = 0.232 m

Since sugar is a molecular substance, the Van't Hoff factor is 1

The normal boiling point of pure ethanol is 78.37 °C

ΔT = K m i

ΔT = 1.22 °C⋅kg/mol * 0.232 m * 1

ΔT = 0.28°C

Boiling point of solution =  78.37 °C + 0.28°C = 78.65°C

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Answer:

no

Explanation:

no its not legal to purchase it

Answer:

(O)oxygen

Explanation:

hope it helps

Answer:

Using temperature or a thermometer

Explanation:

Since AKE equals = temperature, you can find the temperature of a substance with a thermometer, which gives the temperature.

Answer:

Using a temperature scale

Answer:

.265 mol

Explanation:

We want to find the number of moles in 15.9 grams of SiO2

First we need to find the molar mass of SiO2

Looking at a periodic table 1 atom of Si has a mass of 28.086g

And 1 atom of oxygen has a mass of 15.999g

So the molar mass of SiO2 would be 28.086(1) + 15.999(2) = 60.084g/mol

Now we can use dimensional analysis to find the number of moles in 15.9 grams of SiO2

Setting up the table we get

[tex]\frac{15.9gSiO_2}{} *\frac{1 molSiO_2}{60.084gSiO_2}[/tex]

the g SiO2 cancels out and we get

[tex]\frac{15.9}{} *\frac{1 molSiO_2}{60.084}[/tex]

we then evaluate 15.9 * 1/60.084 to get .265 mol

So there are .265 moles in 15.9 grams of SiO2

Answer:

Hence mass of 2 moles of NaCl is 117g. The molar mass of NaCl is 58.44 g/mol. To calculate mass in grams, multiply the given moles by the molar mass.

Answer:

etermine the molarity for a solution of 29.2 g of NaCl is dissolved into 0.50 L of water ... - did not match any news results.

Explanation:

Answer:

One Factor Is Ionization Energy

Another Factor Is The Electronegativity Of The Element

K is the most reactive metal from the given as it is more electronegativity..... and the valence electron are close to the nucleus...

Explanation:

Hope it helps!!!

Answer:

Potassium (K)

Explanation:

It has the furthest distance from the nucleus and the outer most electron shell meaning that the forces between the two are weaker which as a result makes it easier for Potassium to lose it's outer electron.

The amount of substance left after n half cycle of caesium is given by -

[tex]\green{ \underline { \boxed{ \sf{N_t= \frac{N_o}{2^n}}}}}[/tex]

where

[tex]\green{ \underline { \boxed{ \sf{Number \: of \: half \:cycle = \frac{Given\:Time \: period}{Period \: of \: half \:cycle}}}}}[/tex]

[tex]\begin{gathered}\\\implies\quad \sf Number \: of \: half \:cycle = \frac{90.9}{30.3}\\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf Number \: of \: half \:cycle = 3 \\\end{gathered} [/tex]

Now,

[tex]\begin{gathered}\\\implies\quad \sf N_t= \frac{N_o}{2^n} \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf N_t= \frac{60}{2^3} \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf N_t= \frac{\cancel{60}}{\cancel{8}} \\\end{gathered} [/tex]

[tex]\begin{gathered}\\\implies\quad \sf N_t= 12.5 g \\\end{gathered} [/tex]

Therefore, 12.5 gram of cesium would be left over 90.9 years.

Answer:

n = 0.157

Explanation:

To solve that question, we need to follow this equation: PV = nRT

V = 435ml (must be convert to L by ÷ by 1000) = 0.435 L

P = 6780 torr (must be convert to atm by ÷ by 760) = 8.92 atm

T = 28C (must convert to K by + 273.15) = 301.15 K

R = 0.08205

n = ?

PV = nRT

8.92 x 0.435 = n x 0.08205 x 301.15

3.88 = n x 24.7

Divided both sides by 24.7 (left n)

3.88/24.7 = n

0.157 = n

So 0.157 is the answer!

Answer:

contact force

Explanation:

I have no idea why

ΔG for the formation of H₂O is -228.6 kJ.

Gibb's free energy of the reaction is calculated as:

ΔG = G for product - G for reactant

Given chemical reaction is:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

In the question, given that:

ΔG for the reaction = -957. 9 kJ

ΔGf of NH₃ = -16. 66 kJ/mol

ΔGf of NO = 86. 71 kJ/mol

Equation for ΔG will be written as:

ΔG = (ΔGf of NO + ΔGf of H₂O) - (ΔGf of NH₃+ ΔGf of O₂)

ΔGf of O₂ = 0

-957. 9 = (4×86. 71 + 6×ΔGf of H₂O) - (4×-16. 66 + 5×ΔGf of O₂)

-957. 9 = 346.84 + 6ΔGf of H₂O + 66.64

ΔGf of H₂O = (-957. 9 - 346.84 - 66.64) / 6

ΔGf of H₂O = -228.56 kJ ≅ -228.6 kJ

Hence, option (1) is correct i.e. -228.6 kJ is the ΔGf of H₂O.

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Answer:

medium length fur

Explanation:

incomplete dominance is when the two traits kind of combine.  u would have to make a punnet square, but in short, 100% of the offspring would have medium length fur

Answer:

1. The smallest particle of an element is?

B. Atom

2. The smallest particle of matter is?

B.Atom

Explanation:

An atom is the basic building block of chemistry. It is the smallest unit into which matter can be divided without the release of electrically charged particles. It also is the smallest unit of matter that has the characteristic properties of a chemical element.

Answer:

pH = 12.08.

Explanation:

Write the base reaction of methylamine:

[tex]\displaystyle \text{H$_2$O}_\text{($\ell$)} + \text{CH$_3$NH$_2$}_\text{(aq)} \rightleftharpoons \text{CH$_3$NH$_3$}^+_\text{ (aq)} + \text{OH}^-_\text{ (aq)}[/tex]

The equilibrium constant expression will hence be:
[tex]\displaystyle K_b = \frac{\left[\text{CH$_3$NH$_3$}^+\right]\left[\text{OH}^-\right]}{\left[\text{CH$_3$NH$_2$}\right]}[/tex]

As the reaction proceeds, x amounts of CH₃NH₃⁺ and OH⁻ will be formed and (0.35 M - x) amounts of CH₃NH₂ remains.

Assuming that the change to CH₃NH₂ is negligible, we have that:
[tex]\displaystyle \begin{aligned} K_b & = \frac{(x)(x)}{(0.35 -x)} \\ \\ (4.4\times 10^{-4}) & \approx \frac{x^2}{0.35} \\ \\ x & =0.012 \end{aligned}[/tex]

Hence, [OH⁻] = 0.012 M.

Find [H⁺]:

[tex]\displaystyle \begin{aligned} \ [\text{H}^+][\text{OH}^-] & = 1.0\times 10^{-14} \\ \\ [\text{H}^+] (0.012) & = 1.0\times 10^{-14} \\ \\ [\text{H}^+] & = 8.3\times 10^{-13} \text{ M}\end{aligned}[/tex]

Hence, the pH is:
[tex]\displaystyle \begin{aligned} \text{pH} & = -\log [\text{H}^+] \\ \\ & = -\log (8.3\times 10^{-13}) \\ \\ & = 12.08\end{aligned}[/tex]

In conclusion, the pH is about 12.08.

Answer:

did u fr ask what hydrogen is its a chemical element

Explanation:

Answer

Explanation:

he most recent Ebola virus outbreak in West Africa, which was unprecedented in the number of cases and fatalities, geographic distribution, and number of nations affected, highlights the need for safe, effective, and readily available antiviral agents for treatment and prevention of acute Ebola virus (EBOV) disease (EVD) or sequelae1. No antiviral therapeutics have yet received regulatory approval or demonstrated clinical efficacy. Here we report the discovery of a novel small molecule GS-5734, a monophosphoramidate prodrug of an adenosine analogue, with antiviral activity against EBOV. GS-5734 exhibits antiviral activity against multiple variants of EBOV and other filoviruses in cell-based assays. The pharmacologically active nucleoside triphosphate (NTP) is efficiently formed in multiple human cell types incubated with GS-5734 in vitro, and the NTP acts as an alternative substrate and RNA-chain terminator in primer-extension assays using a surrogate respiratory syncytial virus RNA polymerase. Intravenous administration of GS-5734 to nonhuman primates resulted in persistent NTP levels in peripheral blood mononuclear cells (half-life, 14 h) and distribution to sanctuary sites for viral replication including testes, eyes, and brain. In a rhesus monkey model of EVD, once-daily intravenous administration of 10 mg kg−1 GS-5734 for 12 days resulted in profound suppression of EBOV replication and protected 100% of EBOV-infected animals against lethal disease, ameliorating clinical disease signs and pathophysiological markers, even when treatments were initiated three days after virus exposure when systemic viral RNA was detected in two out of six treated animals. These results show the first substantive post-exposure protection by a small-molecule antiviral compound against EBOV in nonhuman primates. The broad-spectrum antiviral activity of GS-5734 in vitro against other pathogenic RNA viruses, including filoviruses, arenaviruses, and coronaviruses, suggests the potential for wider medical use. GS-5734 is amenable to large-scale manufacturing, and clinical studies investigating the drug safety and pharmacokinetics are ongoing.

Main

The 2013–2016 outbreak of EVD in West Africa was the largest and most complex EBOV outbreak in the recorded history of the disease, with >28,000 EVD cases and >11,000 reported deaths1. Medical infrastructures in Guinea, Sierra Leone, and Liberia were seriously impacted by a loss of >500 healthcare workers1. Additionally, EVD-related sequelae (joint and muscle pain, as well as neurological, ophthalmic, and other symptoms) together with viral persistence and recrudescence in individuals who survived the acute disease have been documented2,3,4,5.

EBOV is a single-stranded negative-sense non-segmented RNA virus from the Filoviridae family. In addition to EBOV, other related viruses, namely Marburg, Sudan, and Bundibugyo viruses, have caused outbreaks with high fatality rates6. Although the efficacy of various experimental small molecules and biologics have been assessed in EVD animal models and in multiple clinical trials during the West African outbreak7,8,9,10,11,12,13,14,15,16,17,18, there are no therapeutics for which clinical efficacy and safety have been established for treatment of acute EVD or its sequelae. The availability of broadly effective antiviral(s) with a favourable benefit/risk profile would address a serious unmet medical need for the treatment of EBOV infection.

A 1′-cyano-substituted adenine C-nucleoside ribose analogue (Nuc) exhibits antiviral activity against a number of RNA viruses19. The mechanism of action of Nuc requires intracellular anabolism to the active triphosphate metabolite (NTP), which is expected to interfere with the activity of viral RNA-dependent RNA-polymerases (RdRp). Structurally, the 1′-cyano group provides potency and selectivity towards viral RNA polymerases, but because of slow first phosphorylation kinetics, modification of parent nucleosides with monophosphate promoieties has the potential to greatly enhance intracellular NTP concentrations20. GS-5734, the single Sp isomer of the 2-ethylbutyl L-alaninate phosphoramidate prodrug (Supplementary Information), effectively bypasses the rate-limiting first phosphorylation step of the Nuc (Fig. 1a). In human monocyte-derived macrophages, incubation with GS-5734 caused rapid loading of cells with high levels of NTP that persist with a half-life (t1/2) of 24 h following removal of GS-5734 (Extended Data Fig. 1a), resulting in up to 30-fold higher levels compared to incubation with Nuc (Fig. 1b). In cell-based assays, GS-5734 is active against a broad range of filoviruses including Marburg virus and several variants of EBOV (Fig. 1c). GS-5734 inhibits EBOV replication in multiple relevant human cell types including primary macrophages and human endothelial cells with half-maxi

Answer & Explanation:

Glass Condensation

Heat is transmitted from the hot air to the cold glass when it comes into contact with the cold glass. The water vapor near the glass loses energy due to the loss of heat in the surrounding air. Water vapor condenses into liquid on the glass as energy is lost.

Answer:

45.2%

Explanation:

To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100

First lets find the molar mass of Fluoride

Looking at the periodic table Fluoride has a molecular mass of 18.998 g

Now we need to find the molecular mass of NaF

Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g

Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF

Mass of Fluoride = 18.998g

Mass of Sodium Fluoride = 41.988g

Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%

Answer:

Reactant B.

Explanation:

The mass of product that can be formed from reactant B is 33.2 is less than reactant A's (45.6).

Therefore, reactant B is the limiting reactant.

Answer:

heat to kinetic i think. If its wrong please dont report me :)

Explanation:

Answer:

Chemical to heat

Explanation:

The chemicals speed up causing heat.

Lead is more ductile than iron. The correct option is A. Lead is more ductile than iron

From the question, we are to determine the statement which is true

First, we will define the term Young's modulus

Young's modulus is a measure of the stiffness of an elastic material. It is also given as the ratio of stress to strain. That is,

[tex]Young's\ modulus = \frac{Tensile\ stress}{Tensile\ strain}[/tex]

The greater the modulus, the stiffer the material.

Materials with low Young's modulus tend to be ductile and materials with high Young's modulus tend to be brittle

Since the Young's modulus value for iron is higher than that of lead, then iron is stiffer than lead.

Thus, lead is more ductile than iron. The correct option is A. Lead is more ductile than iron

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Answer:

B

Explanation:

took the test hope it help :)

Arrhenius :

acid: a compound that releases H⁺ ions in water

base: a compound that releases OH⁻ ions in water

H2S Arrhenius acid

RbOH Arrhenius base.

Mg(OH)2  Arrhenius base.

H3PO4 Arrhenius acid

The major distinction between emission and absorption spectra is that an emission spectrum has various colored lines, whereas an absorption spectrum contains dark-colored lines. When electrons return to their original energy levels, this is called emission. When electrons absorb energy, they leap to higher energy levels. When ground-state atoms absorb energy from a radiation source, atomic absorption spectra are created. When neutral atoms in an excited state return to the ground state or a lower-energy state, they release energy, resulting in atomic emission spectra.