Answers
The answer ( 13 and 8 )
x²=5²+12²
x²=25+144
x²=169
x=13
x²=5²+6²
x²=25+36
x²=61
x=7.8
x=8
Related Questions
You decide to share some sugar sweetened fruit juice with your friend. You divide the fruit equally into two glasses. Then an additional equal volume of water is added to glass 1. Which glass would have the lower molarity?
Answers
Answer:
Glass 1
Explanation:
Molarity is measured in moles of substance per liter.
For the sake of calculations, let's say that each glass contains 1 mole of juice and 1 liter after it is divided between the glasses. If you add an equal amount of water to glass 1 ( another liter), you now have:
Glass 1 = 1 mole / 2 liters = 0.5 M
Glass 1 = 1 mole/ 1 liter = 1 M
So glass 1 will have a lower molarity
Group the elements into pairs that would most likely exhibit similar chemical properties. It does not matter which pair of elements is pair 1, pair 2, or pair 3, so long as the correct elements are paired.Pair 1 Pair 2 Pair 3 Answer Bank Mg St Kr Ne+
As P
Answers
Answer: Pair 1 has Mg and Sr, Pair 2 has Kr and Ne, Pair 3 has As and P.
Explanation:
A periodic table is a group of elements presented in a tabular form where elements are arranged in a series of 7 rows and 18 columns.
The vertical columns are known as groups and horizontal rows are known as periods.
The elements having similar chemical properties are arranged in one group.
Magnesium (Mg) is the 12th element of periodic table placed at Group 2 and Period 3
Strontium (Sr) is the 38th element of periodic table placed at Group 2 and Period 5
Krypton (Kr) is the 36th element of periodic table placed at Group 18 and Period 4
Neon (Ne) is the 10th element of periodic table placed at Group 18 and Period 2
Arsenic (As) is the 33rd element of periodic table placed at Group 15 and Period 4
Phosphorus (P) is the 15th element of periodic table placed at Group 15 and Period 3
As magnesium and strontium are present in the same group, they will have similar chemical properties. Similarly, krypton and neon will form the second pair. Likewise, arsenic and phosphorus will form a pair.
Hence, Pair 1 has Mg and Sr, Pair 2 has Kr and Ne, Pair 3 has As and P.
Organic compounds undergo a variety of different reactions, including substitution, addition, elimination, and rearrangement. An atom or a group of atoms in a molecule is replaced by another atom or a group of atoms in a substitution reaction. In an addition reaction, two molecules combine to yield a single molecule. Addition reactions occur at double or triple bonds. An elimination reaction can be thought of as the reverse of an addition reaction. It involves the removal of two atoms or groups from a molecule. A rearrangement reaction occurs when bonds in the molecule are broken and new bonds are formed, converting it to its isomer. Classify the following characteristics of the organic reactions according to the type of organic reaction.
a. Reactions involving the replacement of one atom or group of atoms.
b. Reactions involving removal of two atoms or groups from a molecule.
c. Products show increased bond order between two adjacent atoms.
d. Reactant requires presence of a π bond.
e. Product is the structural isomer of the reactant.
1. Substitution reaction
2. Addition reaction
3. Elimination reaction
4. Rearrangement reaction
Answers
Answer:
Reactions involving the replacement of one atom or group of atoms. - Substitution reaction
Reactions involving removal of two atoms or groups from a molecule - Elimination reaction
Products show increased bond order between two adjacent atoms - Elimination reaction
Reactant requires presence of a π bond - Addition reaction
Product is the structural isomer of the reactant - Rearrangement reaction
Explanation:
When an atom or a group of atoms is replaced by another in a reaction, then such is a substitution reaction. A typical example is the halogenation of alkanes.
A reaction involving the removal of two atoms or groups from a molecule resulting in increased bond order of products is called an elimination reaction. A typical example of such is dehydrohalogenation of alkyl halides.
Any reaction that involves a pi bond is an addition reaction because a molecule is added across the pi bond. A typical example is hydrogenation of alkenes.
Rearrangement reactions yield isomers of a molecule. Rearrangement may involve alkyl or hydride shifts in molecules.
Reactions involving the replacement of one atom or group of atoms is substitution reaction, reactions involving removal of two atoms or groups from a molecule and products show increased bond order between two adjacent atoms is elimination reaction, reactant requires presence of a π bond in addition reaction and product is the structural isomer of the reactant is rearrangement reaction.
What is chemical reaction?Chemical reactions are those reactions in which reactants undergoes through a variety of changes for the formation of new product.
Substitution reaction: In this reaction any atom or molecule of reactant is replaced by any outside atom or molecule.Addition reaction: In this reaction addition of any reagent takes place across the double or triple bond of any reactant for the formation of product.Elimination reaction: In this reaction any molecule or two atoms will eliminate from the reactant as a result of which we get a bond order increased product.Rearrangement reaction: In this reaction atoms or bonds of a reactant get rearranged for the formation of new product.Hence, classification of above points are done according to their characteristics.
To know more about chemical reactions, visit the below link:
https://brainly.com/question/26018275
how many moles of neon gas have a volume of 0.84 L and a pressure of 4.6 atm at 222k
Answers
Answer:
n = 0.21 moles
Explanation:
Given that,
Volume, V = 0.84 L
Pressure, P = 4.6 atm
T = 222 K
We need to find the number of moles of Neon gas. We know that,
PV = nRT
Where
n is the number of moles
R i the gas constant, R = 0.08206 L-atm/mol-K
Put all the values,
[tex]n=\dfrac{PV}{RT}\\\\n=\dfrac{4.6\times 0.84}{0.08206 \times 222}\\\\n=0.21\ \text{moles}[/tex]
So, there are 0.21 moles of Neon gas.
study the reaction given below in which excess magnesium ribbon (Mg)reacts with 50cm of a diluted sulphuric acid solution at room temperature
Questions
what Changes can be made to the following substance to increase the rate of reaction?
5.1.1 Magnesium
5.1.2 Sulphuric acid
Answers
Answer:
Magnesium reacts with dilute hydrochloric acid in a conical flask which is ... One student can add the magnesium ribbon to the acid and stopper the flask, ... 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.
There are four different starting molecules that one might use to synthesize the illustrated alkyl halide as the major product using an electrophilic addition reaction. Please draw all four of them.
Answers
Answer:
Explanation:
An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.
In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.
The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.
Predict the products from theses reaction, and balance the equations. Include phase symbols.
Reaction : K(s)+Cl2(g)⟶
Reaction :Cu(s)+O2(g)⟶
Answers
Answer:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
Explanation:
Both reactions are synthesis reactions (two substances combine to form another).
Reaction: K(s) + Cl₂(g) ⟶
The product is the binary salt KCl. The balanced chemical equation is:
2 K(s) + Cl₂(g) ⟶ 2 KCl(s)
Reaction: Cu(s) + O₂(g) ⟶
The most likely product is the metal oxide CuO. The balanced chemical equation is:
2 Cu(s) + O₂(g) ⟶ 2 CuO(s)
At 35°C, K = 1.6 × 10^-5 for the reaction
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
Calculate the concentrations of all species at equilibrium for each of the following original mixtures.
a. 2.0 mol pure NOCl in a 2.0 L flask
b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask
Answers
Answer:
a) [NOCl] = 0.968 M
[NO] = 0.032M
[Cl²] = 0.016M
b) [NOCl] = 1.992M
[NO] = 0.008 M
[Cl2] = 1.004 M
Explanation:
Step 1: Data given
Temperature = 35°C = 308K
K = 1.6 × 10^-5
Step 2: The reaction
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2
Step 3
a. 2.0 mol pure NOCl in a 2.0 L flask
Concentration at the start:
Concentration = mol / volume
[NOCl] = mol / volume
[NOCl] = 2.0 / 2.0 L
[NOCl] = 1.0 M
[NO] = 0 M
[Cl] = 0M
Concentration at the equillibrium
[NOCl] = 1.0M - 2x
[NO] = 2x
[Cl2]= x
K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5
1.6*10^-5 = ((2x)² * x) / (1.0-2x)²
x = 0.016
[NOCl] = 1.0 - 2*0.016 = 0.968 M
[NO] = 2*0.016 = 0.032M
[Cl²] = 0.016M
b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask
Concentration at the equillibrium
[NOCl] = 2.0 mol / 1.0 L = 2.0 M
[NO] = 0 M
[Cl2]= 1.0 mol / 1.0 L = 1.0 M
Concentration at the equillibrium
[NOCl] = 2.0M - 2x
[NO] = 2x
[Cl2]= 1.0 + x
K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5
1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)
1.6 *10^-5= (2x)² * 1 )/2.0²
1.6 *10^-5= 4x² / 4 = x²
x = [tex]\sqrt{1.6 *10^-5}[/tex] = 4.0*10^-3
[NOCl] = 2.0 - 2*0.004 = 1.992M
[NO] = 2*0.004 = 0.008 M
[Cl2] = 1+ 0.004M = 1.004 M
Which best expresses the uncertainty of the measurement 32.23 cm?
A.) ±0.05 cm
B.) 0.1 cm
C.) 1%
D.) ±0.01 cm?
Answers
Answer:
D.) ±0.01 cm?
Explanation:
Since 32.23 cm has two decimal places, the uncertainty is taken as one-half the last decimal pace.
The last decimal place is 0.03. Half of this is 0.03 cm/2 = 0.015 cm.
Since we cannot go below two decimal places, we ignore the 5 in 0.015 cm.
So, we have our uncertainty as 0.01 cm.
So, the best expression of the uncertainty in the measurement 32.23 cm is ± 0.01 cm.
So, the answer is D. which is ± 0.01 cm.
Identify the isoelectronic elements.
i. Cl-, F-, Br-, I-, At-
ii. Ne, Ar, Kr, Xe, He
iii. N3-, S2-, Br-, Cs+, Sr2+
iv. N3-, O2-, F-, Na+, Mg2+
v. Li+, Na+, K+, Rb+,Cs+
Answers
Answer:
iv. N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺
Explanation:
Isoelectronic elements are those that have the same number of electrons. So, if at least 2 elements differ in their number of electrons, the series is not of isoelectronic elements.
To know the number of electrons we will consider the atomic number and add electrons if it is an anion and subtract electrons it is a cation.
Identify the isoelectronic elements.
i. Cl⁻, F⁻, Br⁻, I⁻, At⁻. NO. Cl⁻ has 18 electrons (17+1) and F⁻ has 10 electrons (9+1). ii. Ne, Ar, Kr, Xe, He. NO. Ne has 10 electrons and Ar has 18. iii. N³⁻, S²⁻, Br⁻, Cs⁺, Sr²⁺. NO. N³⁻ has 10 electrons (7+3) and S²⁻ has 18 (16+2).iv. N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺. YES. They all have 10 electrons v. Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺. NO. Li⁺ has 2 electrons (3-1) and Na⁺ has 10 (11-1).A chemistry student needs 15.0 g of methanol for an experiment. She has available 320. g of 44.4% w/w solution of methanol in water. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.
Answers
Answer:
33.8 g Solution
Explanation:
A chemistry student needs 15.0 g of methanol for an experiment. The concentration of ethanol in the solution is 44.4% w/w, that is, there are 44.4 g of methanol every 100 g of solution. The mass of solution that would contain 15.0 g of methanol is:
15.0 g Methanol × 100 g Solution/44.4 g Methanol = 33.8 g Solution
Since 33.8 g are required and 320. g are available, there is enough solution for the requirements.
what would be the mass of 44.8 L of CO2 gas at STP?
show work if possible
Answers
Answer:
[tex]m=88.02g[/tex]
Explanation:
Hello there!
In this case, for this ideal gas law problem, it turns out necessary for us to remember that one mole of any gas is contained in 22.4 L at STP and therefore, we can use the following ratio to calculate the moles in 44.8 L of CO2:
[tex]\frac{1mol}{22.4L} =\frac{x}{44.8L}\\\\x= \frac{1mol*44.8L}{22.4L}=2mol[/tex]
Finally, since the molar mass of CO2 is 44.01 g/mol, we calculate the mass as follows:
[tex]m=2mol*\frac{44.01g}{1mol}\\\\m=88.02g[/tex]
Regards!
The combustion of ethylene proceeds by the reaction
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (g)
When the rate of disappearance of C2H4 is 0.13 M s-1, the rate of appearance of CO2 is ________ M s-1.
Answers
Compare the solubility of calcium sulfite in each of the following aqueous solutions:
a. 0.10 M Ca(CH3COO)2
b. 0.10 M K2SO3
c. 0.10 M NaNO3
d. 0.10 M KCH3COO
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.
Answers
Answer:
0.10 M Ca(CH3COO)2- Less soluble than in pure water.
0.10 M K2SO3- Less soluble than in pure water.
0.10 M NaNO3 - More soluble than in pure water.
0.10 M KCH3COO- Similar solubility as in pure water.
Explanation:
We have to cast our minds back to the idea of common ions effect. If any ion is already present in solution, the presence of that ion in solution prevents any solute containing a common ion with the solution from dissolving in that solution. In order words, the presence of a common ion makes a solute less soluble in a solvent than it is in pure water.
For instance, 0.10 M Ca(CH3COO)2 and K2SO3 both contain Ca^2+ and SO3^2- ions respectively which are also contained in the solute calcium sulfite.
The presence of these common ions in solution makes calcium sulfite less soluble in these solutions than it is in pure water because the equilibrium position for the dissolution of the solute lies towards the left hand side.
However, calcium sulfite is more soluble in 0.10 M NaNO3 than in pure water due to displacement reaction between the ions in solution.
The solubility of calcium sulfite and 0.10 M KCH3COO in pure water is quite comparable.
A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction takes place:
2CO(g) + 2NO(g 2Co2(g) +N2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
Answers
Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
Where V is volume and n moles of 1, initial state and 2, final state of the gas
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
The volume of the sample is 17.4LAmmonium sulfate (NH4)2SO4 is made by reacting 25.0 L of 3.0 mol/L H2SO4 with 3.1× 103 L of NH3 at a pressure of 0.68 atm and a temperature of 298 K according to the following reaction .
NH3(g) + H2SO4(aq) → (NH4)2SO4 (aq)
How many grams of ammonium sulfate are produced?
Answers
Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Explanation:
For [tex]H_2SO_4[/tex]:Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
Molarity of [tex]H_2SO_4[/tex] = 3.0 M
Volume of solution = 25.0 L
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]
For [tex]NH_3[/tex]:The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where,
P = pressure of the gas = 0.68 atm
V = volume of gas = [tex]3.1\times 10^3L[/tex]
n = number of moles of gas = ? moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the gas = 298 K
Putting values in equation 2, we get:
[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]
For the given chemical equation:
[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]
By stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]
As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent
Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]
Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
When an electron moves up to higher energy levels, the atom Choose... a photon of light whereas the atom Choose... a photon of light when an electron drops to a lower energy level. The photons emitted from an atom appear as
Answers
Answer:
Explanation:
When an electron moves from a lower energy level to a higher energy level, energy is absorbed by the atom. When an electron moves from a higher to a lower energy level, energy is released and photon is emitted.
this emitted photon is depicted as a small wave-packet being expelled by the atom in a well-defined direction.
Which of the following is considered a standard unit of length in the United States?
O square inch
O acre
O cubic yard
O yard
Answers
Answer:
Yard . I hope this helped:))
describe how lyophobic sols are synthesize by dispersion method
Answers
Explanation:
For preparing lyophobic sol, the substance in bulk is broken down into particles of colloidal dimensions (Dispersion) or aggregating smaller particles into particles of colloidal dimensions (condensation).
Trộn 100ml dung dịch H2SO4 0,03M với 200ml dung dịch HCl 0,03M và 0,001mol Ba(OH)2 0,05M . Hãy tính pH của dung dịch này?
Answers
Answer:
pH = 1.92Explanation:
[H+] = 0.1x0.03x2 + 0.2x0.03 = 0.012 mol
[OH-] = 0.001x0.05x2 = 0.0001 mol
=> [H+] dư = 0.012 - 0.0001 =0.0119 mol
pH = -log[H+] = 1.92
What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87m (subject is science) pls answer fast
Answers
Answer:
43.5
Explanation:
Hope that helps
Calculate the concentration of a solution with 0.8g of NaCl in 280mL of water.
Answers
Answer: The molarity of NaCl solution is 0.0489 M
Explanation:
Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}}[/tex] .....(1)
We are given:
Given mass of NaCl = 0.8 g
Molar mass of NaCl = 58.44 g/mol
Volume of the solution = 280 mL
Putting values in equation 1, we get:
[tex]\text{Molarity of solution}=\frac{0.8\times 1000}{58.44\times 280}\\\\\text{Molarity of solution}=0.0489M[/tex]
Hence, the molarity of NaCl solution is 0.0489 M
What size volumetric flask would you use to create a 1.00M solution using 166.00 g of KI?
Answers
Answer:
A 1 liter volumetric flask should be used.
Explanation:
First we convert 166.00 g of KI into moles, using its molar mass:
Molar mass of KI = Molar mass of K + Molar mass of I = 166 g/mol
166.00 g ÷ 166 g/mol = 1 mol KIThen we calculate the required volume, using the definition of molarity:
Molarity = moles / litersLiters = moles / molarity
1 mol / 1.00 M = 1 LAll of the different types of electromagnetic radiation (light, x-rays, ultraviolet
radiation, and so on) make up the
atomic spectrum
electromagnetic spectrum.
sunlight
spectral lines,
Answers
Answer:
bleh
Explanation:
What would happen to the Earth's hydrosphere if there were no atmosphere?
Answers
What is the observation of heating of iodine crystals
Answers
Answer:
On heating, the van der Waals dispersion forces existing then will easily break as it has a low boiling point and sublimates into gas. On heating iodine in the test tube, iodine evolves as violet fuming gas.
Explanation:
How do we fix climate change?
The biggest problem of course is conspiracy theorys. Some say it is just a hoxe when really their chidrin or grandchidrin will sufer greatly from it. How do we educate ourselfs better?
Answers
Answer:
Hi so your answer is that to helping fix the climate change you have to : speak up , power your home with renewable engery , reduce water waste , dont waste food , and finally invest energy .
Explanation:
Really hope i helped , have a nice day :)
Answer:
we can reduce air pollution,which is one of the main cause of climate change.Climate change is not a hoxe if it is not attending the upcoming generations will suffer greatly.
An atom that ______ electrons is called a positive ion. A. has 0 B. has 8 C. loses D. gains
Answers
Answer:
Gains
Explanation:
It gets more electrons
Sodium is a highly reactive metal and
chlorine is a toxic gas, but when they
come together the resulting material,
sodium chloride, is essential for life.
Which of the following is true when
sodium and chlorine are brought into
contact with one another?
Answers
Answer:
NaCl
Explanation:
[tex]na + cl > nacl[/tex]
This is also a salt
How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
Answers
Answer:
0.375mL of solution of nalorphine must be injected
Explanation:
A solution of 0.40% (w/v) contains 0.40g of solute (In this case, nalorphine), in 100mL of solution. To obtain 1.5mg of nalorphine = 1.5x10⁻³g of nalorphine are needed:
1.5x10⁻³g * (100mL / 0.40g) =
0.375mL of solution of nalorphine must be injected