You can determine which is which by weighting them
Copper usually weighs heavier than iron, so if you weigh them both on the scale, the heavier one would be the copper cube
Hope this helped!
How many moles of oxygen will occupy a volume of 435 mL at 6780 torr and 28 °C?
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!
Burning hydrogen is an example of a chemical reaction. The reaction is
shown in the diagram. Which of the following sthtements about energy in
this reaction is the most accurate?
Answer:
Energy is absorbed
Explanation:
since 1932 i learn this
lab report solubility edge
Answer:
i'm attaching the report i made
Explanation:
(It does include the chart information)
hope this helps! have a wonderful day :)
The solubility of a solid solute increases with temperature.
What is the effect of temperature on the solubility of a solute?The solubility of a solute is the amount of solute that dissolves in a given volume of solvent at a particular temperature.
The solubility of a solute increases with increase in temperature.
Based on the lab results, it was seen that with increase in temperature, the mass of sugar that dissolves in water increases.
50 mL of water at 2 °C dissolved only 80 g of sugar whereas at 102 °C, 250 g of sugar dissolved.
Therefore, it can be concluded that solubility of a solute increases with temperature.
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What is the percent by mass of NaCl in solution if 4. 5 grams NaCl is present in 500g of solution?
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
a type of force where objects touch each other
Answer:
contact force
Explanation:
I have no idea why
Write the expressions for the equilibrium constants of the reactions below.
a.N2 () + 3H2 () ⇌ 2NH3 ()
b. Br2 () + H2 () ⇌ 2HBr ()
c.CO () + H2O () ⇌ H2 () + CO2 ()
Answer:
Hey Man I dont know the Answer Im really sorry
Explanation:
I took the test though
what is hydrogen
who likes Black pink Dancer
i like lalalisa
Answer:
did u fr ask what hydrogen is its a chemical element
Explanation:
What is another way to measure the average kinetic energy of a substance?
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
how many iron atoms are there in 5.33 mol of iron(iii) chloride
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 to all these questions...
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
Determine the pH of a 0.35 M aqueous solution of CH3NH2 (methylamine). The Kb of methylamine is 4.4 ⋅ 10-4.
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.
The mass of product that can be formed from reactant A is 45.6 g, and the mass of product that can be formed from reactant B is 33.2 g. Is reactant A or B the limiting reactant?
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.
how many molecules of water are there in 8.050 x 10^3 grams of water
Answer:This is approximately 2.69 * 10^26 molecules.
Explanation:
i just know
Using the formula M V1 = M2V2 , how many milliliters of a 2.50 M
hydrochloric acid solution is required to make 100.0 mL of a 0.750 M
solution?
1.88
333mL
300mL
30mL
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.
9. What is a light-year? A a unit of mass B a unit of distance C unit of volume
Please help ! it is timed Please help me I'll give 20 points points please
Answer:
B
Explanation:
took the test hope it help :)
Deadline homework if someone could this I make u a brainlist. Thanks
As the rate of physical weathering increases, the amount of chemical weathering also increases because
Answer:
Hehe!
Explanation:
Because Increased surface area provides more surface for chemical weathering to attack the rock,allowing chemical weathering to speed up.
List two products derived from ethylene
Cesium-137 has a half-life of 30.3 years.If you start with 60g, how much cesium would be left over 90.9 years?
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]\sf N_t = Amount \:left \:after \: n \:half \: cycles[/tex][tex]\sf N_o=Initial \:Amount \:of \: radioactive \: element \:[/tex][tex]\sf n= Number \: of \:half \:cycle [/tex][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.
Does silver change color in the presence of poison.
What are the characteristics of a fibrous root system? (Select all that apply.)
made up of many roots
made up of one large root
made up of thin roots
made up of branched roots
Answer:branched roots and thin roots and many roots
Which of the following substances dissolves most readily in water?
a. CH
c. NH3
b. BaSO4
d. CaCO3
Given that Delta. G for the reaction below is –957. 9 kJ, what is Delta. Gf of H2O? 4NH3(g) 5O2(g) Right arrow. 4NO(g) 6H2O(g) Delta. Gf,NH3 = -16. 66 kJ/mol Delta. Gf,NO = 86. 71 kJ/mol –228. 6 kJ/mol –206. 4 kJ/mol 46. 7 kJ/mol 90. 7 kJ/mol.
ΔG for the formation of H₂O is -228.6 kJ.
How we calculate Gibb's free energy of the reaction?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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1. Destiny is working in the lab with a filter. She has a mixture of Aluminum Phosphate (AlPO4)
and water. She knows if she runs the mixture through her filter she will collect some solid
because
a. AIPO, is soluble in water
b. AIPO, is insoluble in water
c. AIPO, is attracted to the filter
d. Water is soluble in water
A bicyclist decelerates with a force of -350 N. If the cyclist and bicycle have a total mass of 100 kg, what is the acceleration?
2.a) when a bydrocarbon fuel is burnt, one of the main products is slightly acidic gas R Whats the name of gas R?
Answer:
Carbon and hydrogen atoms in the hydrocarbon fuel react with oxygen in an exothermic reaction. carbon dioxide and water are produced
A toothpaste contains sodium fluoride (NaF) What percentage of Fluoride is present.(4cs)
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%
Which of the following are sources of chemical energy?
1. Batteries and food
2.Batteries and toasters
3. Fans and food
4. Food and toasters
The sources of chemical energy from the available options would be batteries and foods.
What is chemical energy?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.
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The reaction below proceeds spontaneously at 298 K. NH3(g) Cl2(g)Right arrow. NH4Cl(s) What is the sign of the entropy change, Delta. S? Delta. S is negative because the product is less random than the reactants. Delta. S is negative because the reactants are less random than the product. Delta. S is positive because the product is less random than the reactants. Delta. S is positive because the reactants are less random than the product.
A spontaneous reaction favours the product formation and increases the entropy of the isolated system. The sign of the entropy change will be negative as the product formation is less random.
What is entropy?Entropy is the degree of the disorder or randomness that occurs in the system that lacks available energy and cannot convert the mechanical energy or work.
The entropy of the reaction will be negative as the reactants are present in the gaseous state and the product formed is solid. The product formation was spontaneous and less random.
Therefore, option a. entropy will be negative as product formation is less random.
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Combustion is an example of _____ to _____ energy conversion.
your options are:
chemical
kinetic
heat
nuclear
electrical
gravitational
magnetic
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.