Answer:
a - carbon is not toxic to living things
Explanation:
The reason carbon dating works is that the fresh carbon-14 gets mixed in with the rest of the carbon in the atmosphere and, since it's chemically identical to regular carbon, gets worked into whatever is presently absorbing atmospheric carbon
How many moles are 4.20 * 10 ^ 25 atoms of Ca?
Answer:
~69.744 moles of Ca
Explanation:
Using Avogadro's constant , we know that:
1 mole = 6.022 x 10^23 atoms
S0, the number of moles in 4.20 x 10^25 atoms of Ca:
=(4.20 x 10^25 x 1 )/(6.022 x 10^23)
~69.744 moles of Ca
Q2:How many atoms are in 0.35 moles of oxygen?
1 mole = 6.022 x 10^23 atoms
S0, the number of atoms in 0.35 moles of oxygen:
=[0.35 x (6.022 x 10^23)]
=2.1077 x 10^23 atoms of Oxygen
Hope it helps:)
g (2pts) A 10x transfer buffer solution is 250mM Tris and 1.92M glycine. Buffers are always used at 1x concentration in the lab (unless specified otherwise in the protocol), so we will have to dilute the 10x buffer to 1x before use. What is the concentration of Tris and glycine in the 1x buffer
Answer:
The explanation according to the given question is summarized below.
Explanation:
Given:
Tris,
= 250 mM
Glycine,
= 1.92 M
According to the solution,
For the dilution pf 10X to 1X buffer, we get
= [tex]1 \ ml \ of \ 10X \ buffer +9 \ ml \ of \ distilled \ water[/tex]
= [tex]10[/tex]
i.e.,
⇒ [tex]10X \ to \ 1X=1:10 \ dilution[/tex]
Now,
⇒ [tex]10X (250 \ mM\ Tris \ HCl, 1.92M\ Glycine, and\ 1 \ percent (\frac{w}{v} ) SDS) ---->1X(25 \ mM \ Tris \ HCl,0.193 M\ Glycine, and \ 0.1 \ percent(\frac{w}{v} )SDS)[/tex]
What type of reaction does this model represent?
a. Synthesis
b. Single Replacement
c. Decomposition
d. Double Replacement
Answer:
From the diagram above...
its explicitly show a compound that's dissociating or Decomposing into its Constituents.
Answer ....OPTION C.
Radium has a half-life of 1500 years. How long will it take for 250kg of Radium to decay down to less than 10kg
Answer:
6967 years
Explanation:
The radioactive substance left after a periodical t year can be expressed by using the formula:
[tex]Q(t) = Q_oe^{-kt}[/tex]
here;
[tex]Q_o[/tex] = the radioactive initial value.
We need to understand that provided that the radioactive substance will get reduced to half of the provided initial amount after a periodic time, Then:
the half-life of the radioactive substance left is:
[tex]Q(h) = \dfrac{Q_o}{2}[/tex]
Given that:
the half-life = 1500 years
[tex]\dfrac{Q_o}{2}= Q_o e^{-k\times 1500} \\ \\ \dfrac{Q_o}{2}= Q_o e^{ -1500k}[/tex]
Divide both sides by [tex]Q_o[/tex]
[tex]\dfrac{1}{2} =e^{-1500k}[/tex]
Then, find the natural logarithm of both side;
[tex]\mathtt{In \dfrac{1}{2} = -1500 k}[/tex]
[tex]k = \dfrac{1}{-1500}\mathtt{In}\dfrac{1}{2}[/tex]
k = 0.000462
So, after a particular (t) time, a 250 kg radium sample was reduced to 10 kg;
Then:
[tex]10 = 250 e^{-0.000462t}[/tex]
[tex]0.04 = e^{-0.000462t}[/tex]
From both sides, finding the natural logarithm, we have:
In(0.04) = -0.000462t
[tex]t = \dfrac{In(0.04)}{-0.000462}[/tex]
t = 6967.26
Thus, it will take approximately 6967 years for a 250 kg radium sample to get reduced to 10 kg.
what colour is universal indicator when the solution is nuetral?
PLEASE ANSWER ME FAST
Answer:
Green
Explanation:
pH : < 3 = Red
pH : 3 - 6 = Orange or yellow
pH : 7 = Green
pH : 8 - 11 = Blue
Why does water have a high cohesion? *
Hydrogen bonding
O It is not known
O Dispersion forces
Answer:
Dispersion forces
that is the answer
what source of electrical energy have scientists been unable to use?
Answer:
Fusion has powered the sun for billions of years.Yet despite decades of effort, scientists and engineers have been unable to generate sustained nuclear fusion here on Earth. In fact, it's long been joked that fusion is 50 years away, and will always be.
What volume of O2 at NTP is generated from 7.84 g of
FeSO, (NH4)2SO4.6H2O?
Answer:1.12 g
Explanation:Molar mass of compound FeSO4(NH4)2SO4.6H2O = 392.14 g mol-1
Number of moles of compound in 7.84g sample = 7.84 g/ 392.14 g mol-1
= 0.02 mole
1 mole of sample contain 1 mole of Fe
So, 0.02 mole of sample would contain 0.02 moles of Fe.
Mass of Fe equivalent to 0.02 mole = 0.02 mol x molar mass of Fe
= 0.02 mol x 55.84 g mol-1
= 1.12 g
If 143.56 mL of 0.6653 M ammonium carbonate reacts with 175.37 mL of 0.8732 M chromium(III) sulfate in a double replacement reaction and produces 7.543 g of chromium(III) carbonate, what is the percent yield of the reaction
What mass of Cu(s) is electroplated by running 30.0 A of current through a Cu2 (aq) solution for 4.00 h
Answer:
The correct answer is "142 g".
Explanation:
According to the question,
⇒ [tex]Cu^2+(aq)+2e^ ------> Cu(s)[/tex]
where,
Z = 2
F = 96500c
C = 30 A
t = 4h
= [tex]4\times 60\times 60[/tex]
= [tex]14400 \ sec[/tex]
We know that,
Atomic mass of Cu(M) = 63.5 g/mole
Now,
⇒ [tex]W=\frac{Mct}{ZF}[/tex]
On putting the values, we get
[tex]=\frac{63.5\times 30\times 14400}{2\times 96500}[/tex]
[tex]=\frac{27432000}{193000}[/tex]
[tex]=142 \ g[/tex]
a gas tank contains 15 moles of oxygen and 20 moles helium what is apertial pressure of helium in this mixture
Answer:
D0wnload Phot0Math......................
Explanation:
What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one at 500 K
Answer:
The entropy change in the environment is 3.62x10²⁶.
Explanation:
The entropy change can be calculated using the following equation:
[tex]\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})[/tex]
Where:
Q: is the energy transferred = 5.0 MJ
[tex]k_{B}[/tex]: is the Boltzmann constant = 1.38x10⁻²³ J/K
[tex]T_{i}[/tex]: is the initial temperature = 1000 K
[tex]T_{f}[/tex]: is the final temperature = 500 K
Hence, the entropy change is:
[tex] \Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26} [/tex]
Therefore, the entropy change in the environment is 3.62x10²⁶.
I hope it helps you!
When 229 J of energy is supplied as heat at constant pressure to 3.0mol Ar(g) the temperature of the sample increases by 2.55K. Calculate the molar heat capacities at constant volume and constant pressure of the gas.
Answer:
The answer is "[tex]\bold{21.616\ \ \frac{J}{mol K}}[/tex]".
Explanation:
when it is a constant pressure:
[tex]Cp= \frac{\Delta q}{n\Delta T} \\\\[/tex]
[tex]=\frac{229}{3\times 2.55}\\\\=\frac{229}{7.65}\\\\=29.93 \frac{J}{mol\ K}\\[/tex]
and,
[tex]Cp-Cv=R\ \ (Mayer's\ formula)[/tex]
then,
[tex]Cv=Cp-R\\\\[/tex]
[tex]=29.93-8.314\\\\=21.616\ \ \frac{J}{mol K}[/tex]
Is HNO3 an acid or a base
I NEED HELP ASAP
Answer:
HNO3 is a potent acid, a base, a nitrating agent and a heavy oxidising agent at times. In the presence of a stronger acid, it serves as a base.
Explanation:
Answer:
It is a strong acid
Explanation:
How many molecular orbitals are present in the valence band of a sodium crystal with a mass of 4.14 g
Answer:
Explanation:
Given that:
The mass of sodium crystal = 4.14 g
As a result, the number of atoms in the sodium crystal is:
[tex]=4.14 \g \times \dfrac{ 1 \ mol}{23.0 \ g }\times \dfrac{6.023 \times 10^{23} \ atoms}{1 \ mol}[/tex]
= 1.084 × 10²³ atoms
In the valence orbital of the sodium atom, there is just one electron.
As a result, the number of molecular orbitals produced by these atoms is= 1.084 × 10²³ molecular orbital
Because valence bands are created from occupied molecular orbitals, the number of valence bands created from the aforementioned molecular orbitals is:
[tex]= \dfrac{ 1.084 \times 10^{23}}{2}[/tex]
= 5.420 × 10²² orbitals.
I need help Type th temperature (in Kelvin) in the left column and the volumes in the right column, being sure to keep pairs of data together. then click resize window to fit data.
the data appears to be
linear
quadratic
exponential
logarithmic
Answer:
Linear
Explanation:
You have made a 2.5 molar solution (2.5 moles per liter) of a particular chemical. If the molecular weight of the chemical is 28 g (28 g/mol), how many kg of the chemical are dissolved in 1 L of this solution
Answer:
Mass of solute in Kg = 0.07 Kg
Explanation:
Given
Molecular weight of the chemical = 28 grams per mole
Solution consists of 2.5 moles of solution
Mass of solute in the solution = 2.5 *28 = 70 grams
Mass of solute in Kg = 70 grams/1000 = 0.07 Kg
Brainiest and 10 Points
Which has a HIGHER frequency?
A. Visible
B. Ultraviolet
Answer:
B. Ultraviolet
Explanation:
UV has a higher frequency and shorter wavelength than visible light
If ammonium phosphate reacts with sodium chloride in aqueous solution, what are the products?
Answer:
[tex](NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)[/tex]
Explanation:
Hello there!
In this case, according to the given information, we can set up the appropriate chemical equation when ammonium phosphate reacts with sodium chloride in aqueous solution:
[tex](NH_4)_3PO_4(aq)+NaCl(aq)\rightarrow NH_4Cl(aq)+Na_3PO_4(aq)[/tex]
Which stands for a double replacement reaction, whereby ammonium changes phosphate to chloride and sodium changes chloride to phosphate on the products side. In addition, we can balance the aforementioned equation as shown below:
[tex](NH_4)_3PO_4(aq)+3NaCl(aq)\rightarrow 3NH_4Cl(aq)+Na_3PO_4(aq)[/tex]
Regards!
At what temperature do KCl and NaCl have the same solubility?
Answer:
they have the same molar solubility at 44.5 degrees celsius
(1 point) Which compound produces the greatest number of ions when one mole of it is dissolved in water
The question is incomplete, the complete question is;
Which produces the greatest number of ions when one mole dissolves in water? a. NaCl b. NH4Cl c. NH4NO3 d. Na2SO4
Answer:
Na2SO4
Explanation:
If we consider the compounds listed in the options one after the other;
NaCl produces two moles of ions in solution
NH4Cl produces two moles of ions in solution
NH4NO3 produces two moles of ions in solution
Na2SO4 produces three moles of ions in solution
We can see that Na2SO4 produces the greatest number of ions when one mole of the substance is dissolved in water, hence the answer above.
what is a compound ? Give five examples ?
[tex]\huge\mathsf{\red{\underline{\underline{Compound}}}}[/tex]
[tex]{\green{\dashrightarrow}}[/tex]A chemical compound is a chemical substance that is made of two or more atoms of different elements that share a chemical bond.
[tex]{\green{\dashrightarrow}}[/tex]A chemical formula represents the ratio of atoms per element that make up the chemical compound.
[tex]\large{\pink{\sf{5~ Examples~ of~ Compound~ are:-}}}[/tex]
Example 1 :-Water (H2O, consisting of 2 hydrogen atoms and one oxygen atom)Example 2 :- Carbon dioxide (CO2, consisting of one carbon atom and two oxygen atoms)Example 3 :- Sodium Chloride (NaCl, consisting of one sodium atom and one chloride atom)Example 4:-Methane (CH4, consisting of one carbon atom and four hydrogen atoms)Example 5 :- Pure glucose is a compound made from three elements - carbon, hydrogen, and oxygen. The ratio of hydrogen to carbon and oxygen in glucose is always 2:1:1.
Calculate the mass of oxygen in 30 g of CH NH COOH?
Molar mass of CH2NH2COOH - 75
Given mass of CH2NH2COOH - 30
Moles of CH2NH2COOH = Given mass/ Molar mass
moles of CH2NH2COOH = 30/75 = 0.4 mol
One mole of CH2NH2COOH contains 32 gram of oxygen
0.4 mole of CH2NH2COOH will contain = 0.4 × 32= 12.8 g of oxygen
Answer- the mass of oxygen in 30 g of CH2NH2COOH is 12.8 gram!
The mass of oxygen in 30 g of CH₂NH₂COOH is 12.8 g.
We'll begin by calculating the mass of 1 mole of CH₂NH₂COOH and the mass of oxygen in the compound.
1 mole of CH₂NH₂COOH = 12 + (2×1) + 14 + (2×1) + 12 + 16 + 16 + 1 = 75 g
Mass of oxygen in CH₂NH₂COOH = 16 + 16 = 32 g
Thus,
75 g of CH₂NH₂COOH contains 32 g of Oxygen.
With the above information, we can obtain the mass of oxygen in 30 g of CH₂NH₂COOH. This can be obtained as follow:
75 g of CH₂NH₂COOH contains 32 g of Oxygen.
Therefore,
30 g of CH₂NH₂COOH will contain = (30 × 32) / 75 = 12.8 g of Oxygen.
Thus, 12.8 g of Oxygen is present in 30 g of CH₂NH₂COOH
Complete question:Calculate the mass of oxygen in 30 g of CH₂NH₂COOH.
Learn more about mass composition:
https://brainly.com/question/11617445
25.0 g of HCl is added to 100.0 g of water. The density of the resulting solution is 1.19 g/mL. Calculate the molality of
Answer:
The molality is "6.86 m".
Explanation:
Given:
Mass of of HCL,
= 25 g
Mass of water,
= 100 g
Density,
= 1.19 g/mL
Total mass of solution,
= 125 g
Now,
The number of moles of HCl will be:
= [tex]\frac{25}{36.458}[/tex]
= [tex]0.6857[/tex]
The solution volume will be:
= [tex]\frac{125}{1.19}[/tex]
= [tex]105.04 \ mL[/tex]
hence,
The molality will be:
= [tex]\frac{No. \ of \ moles \ of \ solute}{Mass \ of \ solvent (Kg)}[/tex]
= [tex]\frac{0.6857}{0.1}[/tex]
= [tex]6.857 \ m[/tex]
or,
= [tex]6.86 \ m[/tex]
If 2.45 g of iron are placed in 1,5 L of 0.25M HCl, how many grams of FeCl2 are obtained? Identify the limiting and excess reactants in this single replacement reaction. Fe + 2HCl = FeCl2 + H2
Answer:
5.56g of FeCl2 can be produced
Explanation:
To solve this question we must find the moles of each reactant. With the moles and the chemical equation we can find limiting reactant. With limiting reactant we can find the moles of FeCl2 and its mass as follows:
Moles HCl:
1.5L * (0.25mol / L) = 0.375 moles HCl
Moles Fe -Molar mass: 55.845g/mol-
2.45g * (1mol / 55.845g) = 0.0439 moles Fe
For a complete reaction of 0.375 moles HCl are needed:
0.375 moles HCl * (1mol Fe / 2mol HCl) = 0.1875 moles Fe
As there are just 0.0439 moles Fe, Fe is limiting reactant
1mol of Fe produce 1 mole of FeCl2, 0.0439 moles Fe produce 0.0439 moles of FeCl2. The mass is:
Mass FeCl2 -Molar mass: 126.751g/mol:
0.0439 moles Fe * (126.751g / mol) =
5.56g of FeCl2 can be producedInterpreting Velocity vs. Time Graphs
Velocity vs Time
Use the information presented in the graph to answer
the questions,
Which segments show acceleration?
Which segment indicates that the object is slowing
down?
What is the velocity of segment B?
What is the acceleration of segment B?|
50
B
40
30
Velocity (m/s)
20
А
10
0
8 9 10
1 2 3 4 5 6 7
Time (s)
Answer:first part is A and C, second part is C, third part is 40 m/s and the fourth part is zero
Explanation: I guessed and got them right lol
An archaeologist finds 14C in a sample of material to be decaying at 20 Geiger Counter clicks per second. A modern equivalent sample of the same material decays at 160 clicks per second. The half-life of 14C is 5,730 years. How old is the sample
Answer:
17188 years
Explanation:
Recall the formula;
0.693/t1/2 = 2.303/t log(No/N)
t1/2 = half life of 14C
No = initial activity of 14C
N = activity of 14C at time t
t = age of the sample
Substituting values
0.693/5730= 2.303/t log(160/ 20)
1.21 * 10^-4= 2.0798/t
t = 2.0798/1.21 * 10^-4
t = 17188 years
Which is the same as moving the decimal point 3 places to the left in a decimal
number?
Multiplying the number by 100
Dividing the number by 1,000
Multiplying the number by 1,000
Dividing the number by 100
Answer:
B. Dividing the number by 1,000
Explanation:
Every time you move the decimal ,add a zero.Since the decimal is moved three times ,there are three zeroes. When you move left ,the number is getting smaller so it is division.
Soap chemical name and chemical formula
plsss help
Answer:
the chemical name for soap is a lot, for example there is Sodium Talowate, Sodium Palmate, and even Sodium Cocoate, but there is more
Explanation:
the chemical formula for it is RCOO-Na+ it has 12 - 18 carbon atoms.
What would the IUPAC name be?
Answer:
methyl ethanoate
Explanation:
To name the compound given above, the following must be obtained:
1. Determine the functional group of the compound.
In this case, the functional group is R–COOR' where R and R' are alkyl groups. Thus, the compound is an ester.
2. Determine the longest chain before the functional group and the compound after the functional group.
In this case, the longest chain before the functional group is carbon 2 i.e ethane and the compound after the functional group is methyl.
3. Name the compound by naming the compound after the functional group first, followed by the compound before the functional group and ending it with –oate.
This is illustrated below:
After the functional group => methyl
Before the functional group => ethane
Name of the compound => methyl ethanoate