PLEASE HELP ASAPPPPPP
Answer:
accelerating upward would be your answer
A cars is moving at 12 m/s and has a mass of 600 kg. What is the kinetic energy for the car? (Formula:KE= 1/2 MV2) 36,300J 43,200J 72,600J 86,400J
Explanation:
ano po topic nyo para maayos sagut
Learning task 1: as you start exploring the subject answer the following question write your answer on your answer sheet
need help pls
In this exercise, it is necessary to know the systems that our bodies have and complete the sheet that will correspond to these answers:
1- Digestive
2- Urinary
3- Respiratory
4- Skeletal
5- Muscular
6- Integumentary
7- Cardiovascular
8- Endocrine
9- Nervous
10- Reproductive
So looking in more detail, the number one corresponds to the digestive system because it is the digestive process that starts in the mouth. The second is related to the urinary system, which is responsible for the filtration and elimination of excess water. The respiratory system corresponds to the third option, which is the gas exchange that occurs in the pulmonary alveoli.
The numbers 4, 5, and 6 are very important systems for the support of the human body, being respectively the skeleton system, the muscular system, and the integumentary system. The first system is responsible for support while the second is better known for locomotion and finally the third is known for covering the human body by the dermis.
The cardiovascular system is responsible for getting the blood transported throughout the body and pumped by the heart, so this system is related to number 7. Then the endocrine system is responsible for the hormones and how these hormones acted in the body and are number 8.
The number 9 corresponds to one of the most complex and principal systems, the nervous system, which has neurons as its cells and transmits the nervous impulse in response.
The last treated system is the reproductive system, known for its reproductive gametes, and its most important role is to maintain the reproduction of the species.
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1. Calculate the density of an object that is has a mass of 24 grams and a volume of 6 cm cubed
Answer:
4g/cm cubed
Explanation:
density=mass/volume
d=24/6
d=4
What best describes the transition from gas to liquid
It's condensation in which gas particles have a higher kinetic energy..
0.500 mol aluminium hydroxide, Al(OH)3 reacts with 0.500 mol sulphuric acid, H2SO4 to produce aluminium sulphate and water.
a) Write the balanced equation for the reaction.
b) Which reactant is limiting reactant?
c) How many moles of excess reactant is used in the experiment?
d)Determine how many moles of aluminium sulphates was obtained if the percentage yield of aluminium sulphate during the experiment is 77%.
[9 marks]
a) The balanced equation for the reaction would be as follows:
[tex]2 Al(OH)_3 + 3 H_2SO_4 ---> Al_2(SO_4)_3 + 6 H_2O[/tex]
b) The mole ratio of aluminum hydroxide to sulfuric acid is 2:3. This means that every 1 mole of the aluminum hydroxide would require 1.5 moles of sulfuric acid.
0.5 mole aluminum hydroxide would require:
0.5 x 3/2 = 0.75 moles of sulfuric acid.
But only 0.500 moles of sulfuric acid is present. Thus, the limiting reagent is sulfuric acid.
c) With 0.5 moles sulfuric acid, the mole of aluminum hydroxide required would be:
0.5 x 2/3 = 0.33
Excess moles of aluminum hydroxide = 0.5 - 0.33
= 0.17 moles
d) The mole ratio of sulfuric acid to aluminum sulfate produced is 3:1. With 0.5 moles sulfuric acid, the mole of aluminum sulfate produced would be:
0.5 x 1/3 = 0.17 moles
But the percentage yield is 77%
77/100 x 0.17 = 0.13 moles
Thus, the moles of aluminum sulfate that would be obtained with a percentage yield of 77% would be 0.13 moles.
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A student would like to determine how heating a liquid changes its volume. The student hypothesizes that the liquid will increase in volume. The following list shows the steps taken by the student in order to test the hypothesis.
1.Select the liquid to test.
2.Place the liquid in a sealed container.
3.Use a Bunsen burner to heat the liquid by 10°C.
4.Measure the volume of the liquid.
5.Record the results.
What is wrong with how the student conducted the investigation?
A.
The hypothesis was not valid because it is impossible for liquids to change in volume.
B.
The volume of the liquid should be measured before it is heated.
C.
The student should have increased the temperature of the liquid by more than 10ºC.
D.
The length of time it took for the liquid to be heated should be measured.
Answer:
The volume of the liquid should be measured before it is heated.
Explanation:
Because During an experiment to test how a variable changes a substance, it is important to first observe and record the characteristics of the substance before the variable is introduced. In this case, the variable is heat energy.
If 20.3 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?
2 NO (g) + O₂ (g) → 2 NO₂ (g)
Answer:
0.02 moles of O₂ will be leftover.
Explanation:
Chemistry help!
How much 1.50 M H3PO4, in mL, would you need to add to 50.00 mL of 3.50 M Ca(OH)2 in order to neutralize the solution?
_____ of mL of H3PO4
Answer:
Use the formula below
Explanation:
use the moles ratio for this by writing down the reaction and balancing the equation
The temperature of an object increases by 26.1 °C when it absorbs 3711 J of heat. Calculate the heat capacity of the object.
C = ________________ J/°C
Answer:
[tex]\displaystyle C = 142\text{ } \frac{J}{^\circ C}[/tex]
Explanation:
We can utilize the heat capacity formula:
[tex]\displaystyle q = C\Delta T[/tex]
Where q is the heat energy, C is its heat capacity, and ΔT is the change in temperature.
By substitution, q = 3711 J and ΔT = 26.1 °C. Hence:
[tex]\displaystyle \begin{aligned} (3711 \text{ J}) & = C(26.1\text{ } ^\circ\text{C}) \\ \\ C & = 142\text{ } \frac{J}{^\circ C}\end{aligned}[/tex]
Hence, the heat capacity of the substance is about 142 J per degrees Celsius.
I need help wit this
#1
[tex]\\ \sf\longmapsto2 Ca+O_2\longrightarrow 2CaO[/tex]
Balanced
#2
[tex]\\ \sf\longmapsto N_2+3H_2\longrightarrow 2NH_3[/tex]
Balanced
#3
[tex]\\ \sf\longmapsto 2Cu_2O+C\longrightarrow 4Cu+CO_2[/tex]
Balanced
#4
[tex]\\ \sf\longmapsto H_2O_2\longrightarrow H_2O+O_2[/tex]
Already balanced
calculate the Mass of lead that will be deposited from an aqueous solution of lead(2)salt by the same quantity of electricity that deposit 2.7g of copper from an aqueous solution of copper (2)salt
Answer:
8.82 gm of Pb
Explanation:
Question
calculate the Mass of lead that will be deposited from an aqueous solution of lead(2)salt by the same quantity of electricity that deposit 2.7 g of copper from an aqueous solution of copper (2)saltSince the oxidation# of the Pb and the Cu is the same, we know that the same amount of electricity that will deposit 1 mole of Cu will deposit 1 mole of Pb
2.7 gm of Cu were deposited
the atomic mass of Cu = 63.5
the number of moles of Cu deposited was 2.7/63.5 = 0.0425 moles
so 0.0425 moles of Pb will be deposited.
The atomic mas of Pb is 207.2
0.0425 mole of Pb weigh 0.0425 X 207.2 =
8.82 gm of Pb
Each element can be indentified by the number of _______ found in its nucleus, which also equals the elements _______ _______.
Answer:
Protons, Atomic number
Genes that confer resistance to antibiotics provide a selective advantage to organisms that are exposed to antibiotics. true or false
Answer:
True
Genes providing resistance to antibiotics will provide an organism protection against the antibiotics, which are used to kill it, therefore it is indeed an advantage for their survival
What is the weight of:
i) 1 mole of oxygen atoms?
ii)
i) 3 moles of iron atoms?
3 ?
iii) 20 moles of calcium carbonate?
Answer:
Oxygen = 15.999 g/mol
Iron = 3 × 55.845 = 167.535 g/mol
CaCO3 = 20 × 100.0869 = 2001.738 g/mol
The specific heat of a certain type of cooking oil is 1.75J(g. °C). How much heat energy is needed to raise the temperature of 2.44 kg of this oil from 23. °C to 191 °C?
Answer:
The amount of heat needed is 593.88 kJ.
Um Is it in third period and has three electrons in its highest P orbitals
Answer:
Shut yo musty dusty ah up zaddy
Explanation:
come here big mami
how many chemical bonds are formed in CH4 molecule
Answer:
4 covalent bonds
A solution of HF is titraited with a 0.150M NaOH solution. The pH at the half equivalence point is ? The Ka of HF I 0.00068.
Answer:
don't know00000⁰0000000
To determine the heat of neutralisation, Chelsea placed 50 cm' of sodium hydroxide
solution of concentration 1.0 mol dm- in a polystyrene cup and recorded its
temperature. She then recorded the temperature of 50 cm of sulfuric acid, added it
to the cup, stirred the solution and recorded its maximum temperature, as follows:
. initial temperature of NaOH(aq) = 29.5 °C
. initial temperature of H.SO, (aq) = 29.9 °C
• maximum temperature of the solution = 35.8 °C
Calculate the increase in temperature
2 solutions of NaOH and H₂SO₄ at 29.5 °C are mixed in a coffee-cup calorimeter and after the reaction is completed the temperature is 35.8 °C. The increase in the temperature is 6.3 °C.
To determine the heat of neutralization, Chelsea used a coffee-cup calorimeter.
Initially, she had 2 solutions, NaOH and H₂SO₄, both at 29.5 °C. Upon mixing, the heat was evolved and the final temperature of the solution was 35.8 °C. The neutralization reaction was:
NaOH + H₂SO₄ ⇒ Na₂SO₄ + H₂O
The increase in temperature (ΔT) is equal to the difference between the final temperature and the initial temperature.
[tex]\Delta T = 35.8 \° C - 29.5 \° C = 6.3 \° C[/tex]
This data can be used to calculate the heat of neutralization (q) using the following expression.
[tex]q = c \times m \times \Delta T[/tex]
where,
c is the specific heat capacity of the solutionm is the mass of the solution2 solutions of NaOH and H₂SO₄ at 29.5 °C are mixed in a coffee-cup calorimeter and after the reaction is completed the temperature is 35.8 °C. The increase in the temperature is 6.3 °C.
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Which candle burned the longest time why
Answer:
The one that burned longer because it burned longer
Explanation:
How much is 1 mole of something?
O A. 6.02 x 10
O B. 6.02 x 10-23
O C. 6.02 × 23
O D. 6.02 x 1023
Answer:
see explanation
Explanation:
How much is 1 mole of something?
O A. 6.02 x 10
O B. 6.02 x 10-23
O C. 6.02 × 23
O D. 6.02 x 1023
look carefully at your picture
use a magnifying glass if you need to
I am rewriting the choices the way it appears i your picture
A. 6.02 X 10
B 6.02 X 10^-23
C. 6.02^23
D. 6.02 X 10^23
the answer is D. you left out "to the pow......"
a significant mistake
what is in your picture is 6.02 X 10^23 and means
6.02 X 100,000,000,000,000,000,000,000
what you wrote 6.02 X 1023 means 6158
when we are talking about the number of teensy tiny atoms in a measurable mass we are talking BIGGGGG GIANTTTTT numberswhen written correctly, the answer is D
How is melting simillar evaporation
Answer:
Melting and evaporation both represent changes in matter that involve behavior at the molecular level.
Answer:
well both disappear over time
Explanation:
how many atoms of carbon atoms are in the reactant
Answer:
8
Explanation:
There are 8 carbon atoms in the reactants' side, because you multiply the 2 in front of C4H10 and the 4 under the C atom, and that gives you 8. So, 8 carbon atoms.
Which is not an example of vaporization?
Answer:
boi i need the choices idiot
Explanation:
Nobody can add an answer without context.
But if this helps,
"Vaporization of an element or compound is a phase transition from the liquid phase to vapour. There are two types of vaporization: evaporation and boiling. Evaporation is a surface phenomenon, whereas boiling is a bulk phenomenon." - Wikipedia
https://en.wikipedia.org/wiki/Vaporization
THE POINT (-7,4) IS REFLECTED OVER THE LINE X=-3. THEN, THE RESULTING POINT IS REFLECTED OVER THE LINE Y=X. WHERE IS THE POINT LOCATED AFTER BOTH REFLECTIONS?
Answer: 4,2
The graph should show how to do it.
When the volume of a closed container containing hydrogen gas is increased by a factor of two, how will this affect the number of hydrogen atoms in the container
We have that for the Question "When the volume of a closed container containing hydrogen gas is increased by a factor of two, how will this affect the number of hydrogen atoms in the container"
Answer:
Number of hydrogen atoms depend only on amount of gas which remains constant. Hence, increasing the volume of container has no affect on number of hydrogen atoms.Explanation:
The number of hydrogen atoms will remain the same in the container. Because the container is closed and amount of hydrogen gas remains constant, which means that no extra amount of hydrogen gas is removed or added from the container when volume increases.
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Giving brainly for correct answer! :)
Answer: I think its in order, not really sure. First answer on top. Next answer on the left and the Last answer on the right.
What is the concentration of H+ in a 2.5 M HCl solution?
Answer:
The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
Explanation:
As HCl is a strong acid and hence a strong electrolyte, it will dissociate as
HCl ⟶ H⁺ + Cl⁻
So, The concentration of H⁺ will be 2.5 M (same as HCl)
Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M
-TheUnknownScientist 72
0.500 mol aluminium hydroxide, Al(OH)3 reacts with 0.500 mol sulphuric acid, H2SO4 to produce aluminium sulphate and water.
a) Write the balanced equation for the reaction.
b) Which reactant is limiting reactant?
c) How many moles of excess reactant is used in the experiment?
d)Determine how many moles of aluminium sulphates was obtained if the percentage yield of aluminium sulphate during the experiment is 77%. [9 marks]
The number of moles of excess reactant is used in the experiment is 0.33 moles. The number of moles of aluminium sulphates was obtained 0.129 moles if the percentage yield is 77%.
2Al(OH)3(aq) + 3H2SO4(aq) -----> Al2(SO4)3(aq) + 6H2O(l)
Number of moles of Al(OH)3 = 0.500 mol
Number of moles of H2SO4 = 0.500 mol
We have to determine the limiting reactant as follows;
2 moles of Al(OH)3 reacts with 3 moles of H2SO4
0.500 moles of Al(OH)3 reacts with 0.500 moles × 3 moles/2 moles
= 0.75 moles of H2SO4.
This means that H2SO4 is limiting reactant
Since 3 moles of H2SO4 reacts with 2 moles of Al(OH)3
0.500 moles of H2SO4 reacts with 0.500 × 2 moles /3 moles
= 0.33 moles
Only 0.33 moles of excess reactant was used in the experiment.
3 moles of H2SO4 yields 1 mole of Al2(SO4)3
0.500 moles of H2SO4 yields 0.500 moles × 1 mole/3 moles = 0.167 moles
Theoretical yield of Al2(SO4)3 = 0.167 moles
% yield = actual yield/Theoretical yield × 100/1
Actual yield = % yield × Theoretical yield /100
Actual yield = 77 × 0.167/100
Actual yield = 0.129 moles
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