Answer:
Explanation:
Density is the ratio of mass to volume. The SI unit of density is kg/m³. Given that the mass of the block = 23.4 g = 0.0234 kg
The volume of block = change in water level = volume of water rise after block is placed - volume of water = 28.7 ml - 25 ml = 3.7 ml
Volume of block = 3.7 ml = 0.0037 l
1000 l = 1 m³
0.0037 l = 0.0037/1000 m³ = 3.7 × 10⁻6 m³
Density = mass / volume
Density = 0.0234 kg / 3.7 × 10⁻6 m³ = 6324 kg/m³
An element has three stable isotopes of masses 24, 25, and 26. They exist in abundances of 79%, 10%, and 11%, respectively. What is the average atomic mass, and what is the element? *
Answer:
12 this is your answer
hope you like it
30.0 mL of pure water at 282 K is mixed with 50.0 mL of pure water at 328 K. What is the final temperature of the mixture
Answer:
Explanation:
As the 50mL of water are in a big temperature will give heat to the other sample of water.
With the equation:
Q = m*C*T
Where Q is heat, m is mass of the sample, C is specific heat of the sample (4.184J/molK for pure water) and T is temperature in K
We can determine the energy of each sample of water:
Sample 1:
Q = 30.0g*4.184J/molK*282K
Mass is 30.0g because density of water is 1g/mL
Q = 35396.6J
Sample 2:
Q = 50.0g*4.184J/molK*328K
Q = 68617.6J
Total energy is:
Q = 104014.2J
As total mass is 30+50 = 80g, final temperature is:
104014.2J = 80.0g*4.184J*T
310.7K is final temperature of the mixture
Round off the following number to four significant figures.
19.9743
Answer:
4 sig fis so look at the next number which is "4"
answer is 19.97
all numbers not zeros are significant
can anyone help me with nomenclature?
Explanation:
a chemical nomenclature is a set of rules to generate systematic names for chemical compounds.
A pool is 60.0 m long and 35.0 m wide. How many mL of water are needed to fill the pool to an average depth of 6.35 ft? Enter your answer in scientific notation.
Answer:
4.07 × 10⁹ mL
Explanation:
Step 1: Given data
Length of the pool (L): 60.0 m
Width of the pool (W): 35.0 m
Depth of water in the pool (D): 6.35 ft
Step 2: Convert "D" to m
We will use the relationship 1 m = 3.28 ft
[tex]6.35ft \times \frac{1m}{3.28ft} = 1.94m[/tex]
Step 3: Calculate the volume of water (V)
We will use the following expression.
[tex]V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}[/tex]
Step 4: Convert "V" to mL
We will use the relationship 1 m³ = 10⁶ mL.
[tex]4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL[/tex]
How much is the chemotherapy in FeS2?explain
Answer:
Valence of iron in FeS2
Let the valency of iron be x
x+2(oxidation number of Sulphur)=0 as their is no charge on FeS2
As sulphur belongs to Group VI (chalcogens) so the valency/oxidation number of Sulphur is -2
x+2(-2)=0
x-4=0
x=4
So the valency/oxidation number of iron in FeS2 is +4
Explanation:
Hope it will help you monkey ^_^ it's crab
Answer:
Hope you are better now your parents must be worried about you get well soon drink plenty of milk it will make your bones strong.
NOW TIME FOR WISHES ^_^
What types of knowledge and background did Underwood and Prescott bring to the problem?
Answer:
The answer is on the screenshot
Explanation:
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 45.6 days? Assume that the atomic mass of ³²P is 32.0.
Answer:
The correct answer is 6.65 mg.
Explanation:
Based on the given information, the number of days given is 45.6 days. The original mass of sodium phosphate give is 175 mg. The half-life of phosphorus-32 is 14.3 days, therefore, the n or the number of half life will be,
n = 45.6/14.3 = 3.19
So, after 45.6 days, the mass of sodium phosphate sample left will be,
= Original mass × 1/2ⁿ
= 175 mg × 1/2 ^3.19
= 175 mg × 0.1096
= 34.3 mg of sodium phosphate left after 45.6 days
The molecular mass of Na₃³²PO₄ is 3 (23) + 32 + 4 (16) amu = 165 amu
Therefore, 165 grams of sodium phosphate comprise 32 grams of phosphorus.
So, 34.3 mg or 0.343 gram of sodium phosphate will contain,
= 0.343 × 32/165
= 6.65 mg of P³².
A 0.0678 M solution of a weak base has a pH of 9.44. What is the identity of the weak base? Weak Base Kb Ethylamine (CH3CH2NH2) 4.7 × 10–4 Hydrazine (N2H4) 1.7 × 10–6 Hydroxylamine (NH2OH) 1.1 × 10–8 Pyridine (C5H5N) 1.4 × 10–9 Aniline (C6H5NH2) 4.2 × 10–10
Answer:
The weak base is hydroxylamine
Explanation:
The general equilibrium of a weak base (A), is:
A(aq) + H₂O(l) ⇄ HA⁺ + OH⁻
Where Kb is:
Kb = [HA⁺] [OH⁻] / [A]
As both HA⁺ and OH⁻ comes from the equilibrium of A; HA⁺ = OH⁻. So, you can take:
Kb = [OH⁻]² / [A]
OH⁻ is determined from pH, thus:
pOH = -log [OH⁻]
pOH = 14 - pH
pOH = 14 - 9.44
pOH = 4.56
[OH⁻] = 10^{-4.56}
[OH⁻] = 2.75x10⁻⁵
As the concentration of the weak base is 0.0678M:
Kb = [OH⁻]² / [A]
Kb = [2.75x10⁻⁵]² / [0.0678M]
Kb = 1.1x10⁻⁸ = Kb of Hydroxylamine.
The weak base is hydroxylamineHow many types of ions must be present in solution, at minimum, for the phenomenon of selective precipitation to be observed? Your answer should be a whole number without any decimal places.
Answer:
Two (2) types of ions
Explanation:
Metal ions (cations) and negatively charged ions (anions) of either chloride, carbonate, sulphides or hydroxides are present during the selective precipitation.
Majorly, the process determines the metal ions present by initiating precipitation as certain conditions.
I hope this was helpful.
(5.391 × 10−7) + (2.5531 × 10−6)
Answer:
66.441
I hope this helps!
Answer:
66.441
Explanation:
(5.391 × 10-7) + (2.5531 × 10-6)
(46.91) + (19.531)
66.441
carbon dioxide at 25 degrees celsius and 101.3 kpa has a density of 1.799 kg/m^3 determine gas constant
Answer:
Explanation:
Temperature of gas in absolute scale T = 25 + 273 = 298 .
pressure of gas P = 1.013 x 10⁵ N / m²
density D = 1.799 kg / m³
= 1799 g / m³
From gas formula
PV / T = n R Where P is pressure , V is volume and T is absolute temperature , n is no of moles
P / T = n R / V
P / T = m R /M V where m is mass of gas and M is molecular weight .
m / V = D ( density )
P / T = DR/ M
PM / DT = R
Putting the values
1.013 x 10⁵ x 44 / (1799 x 298)
R = 8.314.09 J / K mole
Based on the results of your titration, what volume of the sample would an adult need to consume to reach the recommended daily amount? 60 mg is the recommened daily amount of vitamin C.
Ascorbic acid Volume of DCPIP in Total Volume of DCPIP
concentration (g/L) final syringe (mL) used (mL)
0.00 2.18 mL 0.18 mL
0.50 3.1 mL 0.6 mL
1.00 4.6 mL 2.2 mL
1.50 5.56 mL 3.39 mL
2.00 6.37 mL 4.37 mL
Unknown Trial 1 3.21 mL 1.21 mL
Unknown Trial 2 3.11 mL 1.11 mL
Unknown Trial 3 3.15 mL 1.06 mL
Answer: 85.7 mL
Explanation:
Given the information from the question as plotted in the graph i will be uploading along side this answer,
Average of total volume of DCPIP used is
= (1.21 + 1.11 + 1.06)mL / 3
= 1.12 mL
and corresponding ( ascorbic acid ) is 0.70 g/L
Two parameter given as volume of DCPIP in final syringe and total volume of DCPIP are quite ambiguous
700mg ⇒ 1 L
THEREFORE volume that contains 60mg = (1000/700) × 60 = 85.7 mL
Given that blood exerts the same osmotic pressure as a 0.15 M NaCl solution, which solution could be the isotonic solution
Answer:
0.15 (M) of NaCl
Explanation:
An isotonic solution is defined as the solution having the same osmolarity, or the same solute concentration, as the another solution. If the two solutions are to be separated by any semipermeable membrane, then water will start flowing in equal parts from each solution and also into the other.
In the context, since it is given that the red blood cells exerts the same osmotic pressure as a 0.15 (M) of NaCl solution.
Thus, 0.15 (M) of NaCl is a isotonic solution.
The central Xe atom in the XeF4 molecule has ________ unbonded electron pair(s) and ________ bonded electron pair(s) in its valence shell.
Answer:
Lewis structure in attachment.
Explanation:
Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an expanded octet.
The central Xe atom in the XeF₄ molecule has two unbonded electron pairs and four bonded electron pairs in its valence shell.
The central Xe atom in the XeF4 molecule has 2 unbonded electron pair(s) and 4 bonded electron pair(s) in its valence shell.
Xenon is the element with the following electronic configuration:
[Kr] 4d¹⁰5s²5p⁶
As we can see there are 6 electrons in the p block of its outer orbitalit means Xenon has 6 bonding electron pairs, therefore the electron geometry of octahedraltwo of the pairs of electrons on the central atom are unbonded, or lone pairs therefore the molecular geometry is square planar.four pairs are bonded in its valence cell.Thus, The central Xe atom in the XeF4 molecule has 2 unbonded electron pair(s) and 4 bonded electron pair(s) in its valence shell.
Learn more:
https://brainly.com/question/18385138
Calculate the density in g/mL of 2.0 L of gasoline that weighs 1.34 kg
Answer:
Density = 0.67 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]Density = \frac{mass}{volume} [/tex]To calculate the Density in g/mL we must first convert the mass from kg to g and the volume from L to mL
For the mass
That's
1.34 kg
1kg = 1000 g
1.34 kg = 1000 × 1.34 = 1340 g
For the volume
That's
2.0 L
1 L = 1000 mL
2.0 L = 2 × 1000 = 2000 mL
So we have
mass = 1340 g
volume = 2000 mL
Substitute the values into the above formula and solve for the Density
That's
[tex]Density = \frac{1340}{2000} \\ = \frac{67}{100} [/tex]We have the final answer as
Density = 0.67 g/mLHope this helps you
Tracy had seen her teacher perform the demonstration many times. The materials for the demonstration were lying on the teacher’s table. Tracy tied her long hair back and used the materials to perform the demonstration on her own. a) Write an action that Tracy did correctly. b) Write an action that Tracy did not follow.
Answer:
Tracy was able to remember what her teacher told her to do and she decided to try it,which was the right thing.
But since it's a demonstration then Tracy is supposed to wear all her protective equipments and with a teacher's permission.I hope this helps.Thanks
Periodic Table High School Reference Sheet
Which of the following equations best describes the energy at point
C, assuming energy is conserved?
A PEA = KEC
B
PEA = KEC - PEC
C. PEB = KEO - PEO
D
PES = PEC
A student graphed distance versus time for an object that moves 14 every 2 s. What is the speed of the object?
Answer:
The speed is 7 every second.
Equation:
14/2 = 7
A student graphed distance versus time for an object that moves 14 every 2 s. The speed of the object is 7 m/sec.
What is speed?The speed is the distance traveled by an object with a specific unit of time. Speed is calculated by dividing distance by time. The SI unit of speed is a meter per second.
S = d / t
Distance is the space or extent between two things. It is measured in meters and kilometers.
Time is a variable that equals the distance divided by time.
The slope of a graph is the rate at which the y and x axis change together, the "rise over the run." Slope can be simply calculated using this equation:
Given, that distance = 14 meters
The time is 2 sec.
By the formula
d = 14 / 2 = 7 m /sec.
Thus, the speed of the object is 7 m/sec.
To learn more about speed, refer to the link:
https://brainly.com/question/7359669
#SPJ2
Approximately how many moles of Al3 are reduced when 0.1 faraday of charge passes through a cell during the production of Al
Answer:
HELLO YOUR QUESTION IS INCOMPLETE ATTACHED BELOW IS THE COMPLETE QUESTION
answer : 0.033 mol ( A )
Explanation:
To determine the number of moles of AI3 that will be reduced when 0.1 Faraday passes through a cell during the production of AI
AI3+ + 3 e -------- AI this means that it will take 3 mol to reduce AI3 to AI
1 mol AI = 3 F electrons
therefore : 1 F electrons = 1/3 mol of AI
therefore : 0.1 F ( Faraday ) = 0.1 * 1/3 = 0.033 mol of AI is needed
Calculate the mass of the air contained in a room that measures 2.50 m × 5.50 m × 3.00 m (density of air = 1.29 g/dm3 at 25°C).
Answer:
Explanation:
Volume of air = 2.5 x 5.5 x 3.00
= 41.25 m³.
Density of air = 1.29 g / dm³
a dm³ = ( 0.1 )³m³
= .001 m³
density in m³ = 1.29 / .001 gm⁻³
= 1290 g / m³
mass of air in the room
= 41.25 x 1290 g
= 53212.5 g
= 53.2125 kg .
Explanation:
please mark me as brainlest
One of the differences between a voltaic cell and an electrolytic cell is that in an electrolytic cell __________.
Answer:
a nonspontanenous reaction is forced to occur
Hope this helps :)
Blake is eliminating contributing factors to ensure accuracy in his results. Which step of the scientific method is he performing?
O Research the hypothesis
O Research the problem
O Test the hypothesis
O Analyze the results
Conclusion
Answer:
C
Explanation:
A sample initially contains 4.4 moles of a radioactive isotope. How much of the sample remains after four half-lives?
Answer:
0.275
Explanation:
Given that:
Initial number of miles of radioactive isotope = 4.4 moles
How much if the sample remains after 4 half life's
Half-life refers to the time taken by a sample to reduce to half of it's initial value.
After the first half life :
Initial / 2 = 4.4 / 2 = 2.2 moles
Second half life :
2.2 / 2 = 1.1 moles
Third half life :
1.1 / 2 = 0.55 Moles
Fourth half life :
0.55 / 2 = 0.275 moles
how many grams of iron(ii) sulfate heptahydrate is needed to prepare 250.0 ml of a 3.25 molar aqueous solution
Answer:
The mass of FeSO₄.7H₂O needed [tex]\simeq[/tex] 225.89 grams
Explanation:
From the given question
The molecular weight of iron(ii) sulfate heptahydrate FeSO₄.7H₂O = 278.02 g/mol
Molarity = 3.25
Volume = 250.0 mL
The mass of FeSO₄.7H₂O needed = [tex]\dfrac{moles \times molecular weight \times 250\ mL}{1000 \ mL}[/tex]
The mass of FeSO₄.7H₂O needed = [tex]\dfrac{3.25 \times 278.02 \times 250}{1000}[/tex]
The mass of FeSO₄.7H₂O needed = 225.89125 grams
The mass of FeSO₄.7H₂O needed [tex]\simeq[/tex] 225.89 grams
The average adult heart pumps about 84./mLs of blood at 72 beats per minute. Suppose you need to calculate how long it would take the average heart to circulate 3600.mL of blood. Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Answer:
42.857 seconds
Explanation:
Given that:
The average adult heart pumps about 84./mLs of blood at 72 beats per minute.
The total volume = 3600 mL
The objective is to calculate how long it would take the average heart circulate 3600 mL of blood.
Since 84 mL of blood takes 1 seconds to pump,
assuming q is the time required for heart to pump 3600 mL of blood
[tex]Time \ ( q )= \dfrac{3600 mL \times 1\ seconds}{84 \ mL}[/tex]
TIme (q) = 42.857 seconds
"Splotchy" most likely refers to a mineral's:
O color
luster
streak
O hardness
Answer:Luster
Explanation: i'm tired so it is luster because splotchy wouldn't be referring to the others so it would be referring to Luster the reflection of light of the object. Hope it helps
PLZ BRAINLILEST I ONLY NEED ONE MORE
Answer:
the answer is color
Explanation:
i got it right on my assignment lol
A 2.0 mL sample of carbon tetrachloride (CCI4 ) masses at 3.16 g. What is the density of this substance in g/mL?
A. 1.5 g/mL
B. 1.58 g/mL
C. 1.6 g/ml
D. 0.63 g/ml
Answer:
The answer is option BExplanation:
The density of a substance can be found by using the formula
[tex]Density = \frac{mass}{volume} [/tex]
From the question
mass of CCl4 = 3.16 g
volume of CCl4 = 2.0 mL
Substitute the values into the above formula and solve for the Density
That's
[tex]Density = \frac{3.16}{2} [/tex]
We have the final answer as
Density = 1.58 g/mLHope this helps you
What is the change in enthalpy when 9.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas
Answer:
-891kj
Explanation:
We have the chemical equation as:
2SO2(g) + O2(g) ----2SO3; ∆H° = 198kj
If sulphur trioxide decomposes we have opposite reaction. Enthalpy of decomposition will not be positive for forward reaction.
2 moles of SO3 ----∆H° = -198Kj
1 mole = -198/2
9 moles = -198/2 x 9
= -891kj
Therefore change in enthalpy when 9 moles is decomposes is -891kj.
You have been injured in the laboratory (cut,burn,etc) first you should
Answer:tell the instructor of the lab
Explanation: