Over long periods of time, the climate of an area can change. Which of these can cause a change in climate?

Answer:

The formula shows that there is one atoms of iron in each formula unit of FeCl3, and by definition the number of molecules or formula units in a mole is Avogadro's Number. Therefore, 5.33 moles contains 5.33 X Avogadro's Number of atoms, which is 3.21 X 1024 atoms, to the justified number of significant digits.

Answer:

The formula shows that there is one atoms of iron in each formula unit of FeCl3 , and by definition the number of molecules or formula units in a mole is Avogadro's Number . Therefore , 5.33 moles contains 5.33 X Avogadro's Number of atoms , which is 3.21 X 1024 atoms , to the justified number of significant digits .

Answer:

did u fr ask what hydrogen is its a chemical element

Explanation:

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.

To know more about Gibb's free energy, visit the below link:

https://brainly.com/question/14415025

The sources of chemical energy from the available options would be batteries and foods.

Chemical energy is a form of energy derived from the chemical properties of materials.

Most batteries contain chemicals that are able to ionize in solution to produce electric currents in circuits. Chemicals in batteries include sodium chloride, nitric acid, sulfuric acid, etc.

Foods contain chemicals that when hydrolyzed in the body, are able to produce energy to sustain the various processes in the body. The chemicals in foods are mostly carbohydrates.

More on chemical energy can be found here: https://brainly.com/question/1371184

Answer:

Hey Man I dont know the Answer Im really sorry

Explanation:

I took the test though

Answer:

B

Explanation:

took the test hope it help :)

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.

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.

Answer:This is approximately 2.69 * 10^26 molecules.

Explanation:

i just know

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:

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:

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

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:

contact force

Explanation:

I have no idea why

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:

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:

(O)oxygen

Explanation:

hope it helps

Answer:30mL

Explanation:

If you add 2.50 with 0.750 divide that by 100 you would get 30.7 something without rounding it would be 30.

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:

A. Atomic number is 8, mass number of M is 16

B. Oxygen, O

C. 8 valence electrons, valency is 2

Explanation:

A. Atomic number = proton number

M2- = 2 extra electrons after forming octet structure, so 10-2=8

8 corresponds to number of protons, so atomic number is 8.

Mass number is number of protons + neutrons = 8+8=16

B. 8 in periodic table corresponds to oxygen

C. Number of valence electrons in an oxide ion is 8, valency is two as it forms -2 charge

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:

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:

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:

Carbon and hydrogen atoms in the hydrocarbon fuel react with oxygen in an exothermic reaction. carbon dioxide and water are produced

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.