Answer:
3.6124 m/kg
Explanation:
Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us 35.4 g of H3PO4 and 100 g of solution to work with.
To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.
Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:
35.4 g x 1 mol / 97.9952 g = 0.3612422 mol
Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.
To find the molality, divide the moles of solute and kilograms of solution.
0.3612422 mol / 0.1 kg = 3.6124 m/kg
the density of gold is 19.3 g/ml. if you had a cube made of pure gold that weighs 58.1 pounds what would be the lengh of a side of this cube
Answer:
11.1 cm
Explanation:
First we convert 58.1 pounds into grams:
58.1 lb * [tex]\frac{453.592g}{1lb}[/tex] = 26353.69 gThen we calculate the volume of the gold cube, using the given density:
26353.69 g ÷ 19.3 g/mL = 1365.48 mL
1365.48 mL is equal to 1365.48 cm³.With the volume of a cube we can calculate the side length:
Length = ∛(1365.48 cm³) = 11.1 cma solution that contains a large amount of solute would be described as what
Answer:
A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.
how many moles of H2o are equivelant to 97.3g h2o
Answer:
Number of mole in H2O = 5.4 moles
Explanation:
Given:
Mass of H2O = 97.3 gram
Find:
Number of mole in H2O
Computation:
We know that molar mass of H2O = 18 g/mole
So,
Number of mole = Given Mass / Molar mass
Number of mole in H2O = Mass of H2O / Molar mass of H2O
Number of mole in H2O = 97.3 / 18
Number of mole in H2O = 5.4055
Number of mole in H2O = 5.4 moles
Which commercial technology commonly uses plasmas?
a radio
a race car
a television
a microwave oven
Answer:
A television is commercial technology commonly uses plasmas.help me plsssssssss.... im timed
Answer:
a
Explanation:
a
Answer:
B. Isotopes of the element.
Explanation:
Isotopes are basically atoms of the same element that contain the same number of protons, but a different number of neutrons.
A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?
Answer:
Mass of sample in mg = 15,285 mg
Explanation:
Given:
Volume of urine sample = 15 ml
Density of sample = 1.019 g/ml
FInd:
Mass of sample in mg
Computation:
Mass = density x volume
Mass of sample in mg = Volume of urine sample x Density of sample
Mass of sample in mg = 1.019 x 15
Mass of sample in mg = 15.285 gram
Mass of sample in mg = 15.285 x 1,000
Mass of sample in mg = 15,285 mg
The element tin has the following number of electrons per shell: 2, 8, 18, 18, 4. Notice that the number of electrons in the outer shell of a tin atom is the same as that for a carbon atom. Therefore, what must be true of tin
The question is incomplete, the complete question is:
The element tin has the following number of electrons per shell: 2.8. 18, 18, 4. Notice that the number of electrons in the outer shell of a tin atom is the same as that for a carbon atom. Therefore, what must be true of tin? Tin is a polar atom and can bind to other polar atoms. Tin has a high molecular weight to give tin-containing molecules greater stabilty. All of the above Tin conform single covalent bonds with other elements, but not double or triple covalent bonds Tincan bind to up to four elements at a time
Answer:
Tin can bind to up to four elements at a time
Explanation:
Certain important points were made in the question about tin and one of them is that tin is an element in the same group as carbon hence it has the same number of valence electrons as carbon.
Carbon is always tetra valent. The tetra valency of carbon is the idea that carbon forms four bonds.
If tin has the same number of valence electrons as carbon, then, tin can bind to up to four elements at a time
Question 13 of 32
The pH of a sample of seawater is 8.0. What is the hydrogen ion concentration of seawater?
A. 8.0 M
B. 1 x 108 M
C. 6.0 M
D. 1 x 10-8 M
Answer:
D
Explanation:
Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 9.32 percent of its original value
Answer:
t(9.32% remaining) = 203 yrs (3 sig. figs.)
Explanation:
All radioactive decay follows a 1st order decay profile. This is defined by the expression ...
A =A₀e^-k·t
A = final activity = 9.32%
A₀ = initial activity = 100%
e = base of natural logs
k = rate constant = 0.693/t(1/2) = (0.693/30.2) yrs⁻¹ = 0.023 yrs⁻¹
t = time of decay = ln(A/A₀)/-k = ln(9.32%/100%)/-0.023 yrs⁻¹
= 203.286637 yrs (calc. ans.)
≅ 203 yrs (3 sig. figs.)
Francine makes several measurements of the mass of a metal block. The data set is shown in the table below.
Measurement
Mass of metal block (g)
1
20.73
2
20.76
3
20.68
4
20.75
After analyzing this data set, Francine calculates a value of 20.73 g.
Which of these characteristics has been calculated?
mean
median
mode
range
Answer:
Mean
Explanation:
The mean of a series of measurements is calculated when all the measurements are added up and then divided by the number of measurements taken, as follows:
Sum of Measurements = 20.73 + 20.76 + 20.68 + 20.75 = 82.92As there are 4 measurements, the mean is:
Mean = 82.92 / 4 = 20.73Answer:
A
Explanation:
Two planets have similar masses but differ in their distances from the Sun.
Which planet experiences a greater gravitational force with the Sun?
A. The planet that is smaller in volume
B. The planet that is larger in volume
C. The planet that is farther from the Sun
D. The planet that is closer to the Sun
Answer:
D. The planet that is closer to the Sun
Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer: The mass of P-32 left from the original sample is 32.07 mg
Explanation:
All radioactive decay processes follow first-order reactions.
Calculating rate constant for first order reaction using half life:
[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)
[tex]t_{1/2}[/tex] = half life period = 14.3 days
k = rate constant = ?
Putting values in equation 1:
[tex]k=\frac{0.693}{14.3days}\\\\k=0.0485days^{-1}[/tex]
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(2)
Given values:
a = initial concentration of reactant = 175 mg
a - x = concentration of reactant left after time 't' = ? mg
t = time period = 35.0 days
Putting values in equation 2:
[tex]0.0485days^{-1}=\frac{2.303}{35.0 days}\log (\frac{175}{a-x})\\\\\log (\frac{175}{a-x})=\frac{0.0485\times 35.0}{2.303}\\\\\log (\frac{175}{a-x})=0.737\\\\\frac{175}{a-x}=10^{0.737}\\\\a-x=\frac{175}{5.457}=32.07mg[/tex]
Hence, the mass of P-32 left from the original sample is 32.07 mg
How many grams of Al2O3 is extracted from 250. g of FeO?
Answer:
[tex]m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Explanation:
Hello there!
In this case, if we consider the following chemical reaction, whereby Al2O3 is produced from Al and FeO:
[tex]3FeO+2Al\rightarrow 3Fe+Al_2O_3[/tex]
Thus, since there is 3:1 mole ratio of FeO to Al2O3, it turns out feasible for us to use their molar masses, 71.844 g/mol and 101.96 g/mol respectively, to obtain the grams of the latter as follows:
[tex]m_{Al_2O_3}=250.gFeO*\frac{1molFeO}{71.844gFeO}*\frac{1molAl_2O_3}{3molFeO} *\frac{101.96gAl_2O_3}{1molAl_2O_3}\\\\m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Regards!
what climate zone is asia?
Answer:
The Tundra Climate
Explanation:
:) hope this helps
Given the balanced reaction: Zn + 2HCl → H2 + ZnCl2
If 5 grams of each reactant are available for the reaction and HCl is known
to be the limiting reactant, which of the following is correct?
O Both reactants will be completely used up.
O There will be excess of both reactants remaining.
O HCl will be completely used up while Zn will remain in excess.
O Zn will be completely used up while HCl will remain in excess.
Answer:
O HCl will be completely used up while Zn will remain in excess.
Explanation:
Zn + 2HCl → H₂ + ZnCl₂In reactions involving two reactants, if one of them is the limiting reactant then the other one has to be the reactant in excess.
Meaning that in this case, the reaction will proceed until HCl is completely used up, and a certain amount of Zn will remain (thus being the reactant in excess).
As a result of the particles in a gas being in constant motion, gas has a _______.
variable volume
variable Pressure
variable Shape
variable mass
Answer:
i think it's variable pressure
if not soo advance sorry :)
Look at picture please
Answer:
Keep temperature constant and increase the pressure of the reaction. The rate of reaction increases.
Explanation:
First of all, the question is asking us to design an experiment to investigate the effect of pressure on the rate of reaction hence the pressure can not be held constant since it is the variable under investigation. This eliminates the first option.
Secondly, increasing the pressure of the reaction means that particles of the gas collide more frequently leading to a greater number of effective collisions and a consequent increase in the rate of reaction according to the collision theory.
Hence the answer above.
1. Complete the following chart for the following atoms:
Element
Mass
number
Number of
electrons
Number of
neutrons
Atomic
Number
(number of
protons)
Potassium
16
56
26
What is the atomic symbol (the nuclide) for the isotope with 15 protons and 16
neutrons? Your answer must be in the form shown:
AX
You must show the correct numbers for A and Z
and have the correct symbol (Z). (3 points)
Explanation:
If there are 15 protons, 15 nuclear particles of unit positive charge, then
Z
=
15
. Now
Z
≡
the atomic number
, and you look at your copy of the Periodic Table, and you find that for
Z
=
15
, the element phosphorus is specified.
But we are not finished. Along with the 15 defining protons, there are also 16 neutrally charged, massive nuclear particles, 16 neutrons, and the protons and neutrons together determine the atomic mass. The isotope is thus
31
P
, which is almost 100% abundant, and an important nucleus for
NMR spectroscopy.
The limiting reactant in a chemical reaction is the reactant __________ Select one: A. for which you have the lowest mass in grams. B. which has the lowest coefficient in the balanced equation. C. which has the lowest molar mass. D. which is left over after the reaction has gone to completion. E. None of the above.
Answer:
i think its A
is it a physical or chemical change when a candle is lit
An aqueous sucrose (C12H22O11) solution must be created for an experiment. If 100.00 mL of 0.200 M solution is needed, what amount of sucrose (in grams) must be weighed out
Answer:
6.85 g
Explanation:
Step 1: Given data
Molar concentration of the solution: 0.200 M (0.200 mol/L)Volume of the solution: 100.00 mL (0.10000 L)Step 2: Calculate the moles of sucrose (solute) required
Molarity is equal to the moles of solute divided by the liters of solution.
M = moles of solute / liters of solution
moles of solute = M × liters of solution
moles of solute = 0.200 mol/L × 0.10000 L = 0.0200 mol
Step 3: Calculate the mass corresponding to 0.0200 moles of sucrose
The molar mass of sucrose is 342.3 g/mol.
0.0200 mol × 342.3 g/mol = 6.85 g
The amount of sucrose (in grams) must be weighed out for the given reaction is 6.846g.
How we calculate mass from moles?Mass of any substance will be calculated by using the moles as:
n = W/M, where
W = required mass
M = molar mass
Given that, molarity of sucrose = 0.2 M
Volume of solution = 100mL = 0.1 L
Relation between moles and molarity is represented as:
M = n/V
On putting values on the above equation we get,
n = (0.2)(0.1) = 0.02 moles
We know that molar mass of sucrose (C₁₂H₂₂O₁₁) = 342.3 g/mole
Now we calculate the required mass by putting values on the first equation as:
W = (0.02)(342.3) = 6.846g
Hence, the required mass of sucrose is 6.846g.
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Why is iodine always Used in a solution containing excess I2
Answer:
If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.
Iodine always used in a solution excess KI is given to aid in the solubilization of free iodine, which would be insoluble in clean water during normal circumstances.
What is Iodine?
Iodine is a kind of element which are mainly used in iodometry titration. It can be represented by I.
What is solution?A solution would be a homogenous mixture of two components, usually a solute as well as a solvent.
Iodimetry would be a technique that uses a standard iodine solution as a titrant for such an oxidizable analyte. When an excessive amount of iodide is used to decrease a chemical while somehow producing iodine.
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An elementary step is defined as a chemical collision in a reaction mechanism. A collection of different types of collisions makes up the reaction mechanism, so elementary steps provide a molecular view of the overall reaction.
a. True
b. False
2KClO3 (s)⇄2KCl (s)+ 3O2 (aq) equilibrium constant
Answer: The equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of reactants each raised to the power of their stoichiometric coefficients.
The concentration of all the solids and liquids are considered to be 1 in the expression of equilibrium constant
For the given chemical equation:
[tex]2KClO_3(s)\rightleftharpoons 2KCl(s)+3O_2(aq)[/tex]
The expression of equilibrium constant follows:
[tex]K_{eq}=[O_2]^3[/tex]
Hence, the equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
describe what it would be like to be an atom
Explanation:
An atom consists of two regions. The first is the tiny atomic nucleus, which is in the center of the atom and contains positively charged particles called protons and neutral, uncharged, particles called neutrons. ... Most atoms contain all three of these types of subatomic particles—protons, electrons, and neutrons.
If 0.5 L of O2(g) reacts with H, to produce 1 L of H2O(g), what is the volume of
H2O(g) obtained from 1 L of O2(g)?
-
0.5 L
2.5 L
2 L
1.5 L
Answer:
2 L
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
From the balanced equation above, we can see clearly that 1 L of O₂ reacted to 2 L of H₂O.
This implies that 2 L of H₂O can be obtained by the reaction of 1 L of O₂.
Thus, option 3 gives the correct answer to the question.
Calculate the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the following balanced chemical equation: 2 Al + 6 HCl → 2 AlCl3 + 3 H2
Answer: The mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
Explanation:
The given balanced reaction equation is as follows.
[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]
Here, the mole ration of Al and hydrogen produced is 2 : 3
As mass of aluminum is given as 26.98 g. So, moles of aluminum (molar mass = 26.98 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol[/tex]
So, when 1 mole of Al reacted then 1.5 moles of hydrogen is produced as per the given mole ratio.
Therefore, mass of hydrogen formed is calculated as follows.
[tex]mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g[/tex]
Thus, we can conclude that the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.
Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3CHO CH3OH CH3CN
Answer:
CH3CH2CH3
Explanation:
Dipole moment is the measure of the polarity of a chemical bond. It is the extent of charge separation in a molecule.
Dipole moment is the product of the magnitude of charge and the distance separating the charges from each other.
The molecule having the lowest dipole moment among the options is the molecule that has the least polarity. The least polar molecule among the options is CH3CH2CH3, it has no polar bonds in its structure.
We have that for the Question "Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3CHO CH3OH CH3CN "
it can be said that
[tex]CH_3CH_2CH_3 have the lowest dipole moment[/tex]From the question we are given
CH3CH2CH3
CH3OCH3
CH3CHO
CH3OH
CH3CN
Generally alkanes have the lowest dipole moment, they have C-H bond which are non polar.
Therefore,
[tex]CH_3CH_2CH_3[/tex] have the lowest dipole moment
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A molecule of composition is replicated in a solution containing unlabeled (not radioactive) GTP, CTP, and TTP plus adenine nucleoside triphosphate with all its phosphorus atoms in the form of the radioactive isotope 32P. Will both daughter molecules be radioactive
Answer:
Please find the complete question in the attached file.
Explanation:
It would only be radioactive if the DNA molecule that employed the poly-T rand as templates. Its other molecule of the daughter would not have been radioactive as it did not need dATP for its replication. While each strand of the second molecule includes t, simultaneous reproduction dATP from both daughter molecules is needed so that each of those is radioactive.
Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04 days. If you begin with 34.7 mg of this isotope, what mass remains after 8.52 days have passed
Answer:
16.6 mg
Explanation:
Step 1: Calculate the rate constant (k) for Iodine-131 decay
We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.
k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹
Step 2: Calculate the mass of iodine after 8.52 days
Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.
ln I = ln I₀ - k × t
ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day
I = 16.6 mg
