8
What happens to solid waste in the circulatory system?
A it's expelled through the lungs
B
It pumps into the coronary circulation
C
It's dropped off in the kidneys
D
It's deposited in the aorta
Answer:c it’s dropped off in the kidneys
Explanation:
I took the quiz
The solid wastes are from the circulatory system is expelled to the kidney where, the nitrogenous wastes like urea and uric acid is excreted as urine from the body.
What is circulatory system ?A circulatory system is an organ system, where the blood is purified and oxygenation of blood takes place. Through circulatory system, the blood reaches throughout the body pumped from the heart through veins.
The organs included in circulatory system are lungs, heart, aorta, veins, blood vessels etc. There are various kinds of blood vessels each having specific functions.
There is a network of blood vessels including arteries and large veins, capillaries that join the venules and other veins. All the nutrients and ions are circulated throughout the body through blood and solid wastes are then expelled to kidney.
Kidney function as a sieve to clean the good fluid from waste products. Uric acid and urea along with water excreated as urine then. Thus, option C is correct.
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A community located downwind from a coal-fired power plant has seen a recent increase in the number of dead and dying trees. A so scientist measured values for the following parameters before and after the trees died off. Which of the following oil data should be used to determine if the coalfired power plant emiations were the cause of the damage to the trees
a. Moisture content and water retention
b. Parent material composition
c. Pesticide and herbicide residue levels
d. Calcium and aluminum levels
Answer:
Option D, Calcium and Aluminum levels
Explanation:
The coal fired power plant releases huge amount of particulate and gaseous emissions such as mercury, sulphur dioxide, nitrogen oxide etc. When there is rain, these gaseous and particulate matter comes to the ground along with rain water and pollute the soil. There are also chances of acid rain due to the presence of sulphur dioxide. Polluted soil and acid rain negatively impact the growth of the plants and causes leaching of Aluminium thereby decreasing the availability of calcium for the plants. Thus, the trees die. Hence, if the amount of Aluminium and Calcium in soil is determined, one can easily deduce the cause of death of trees.
Hence, option D is correct
explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity
What does the cell theory state? Answer F All organisms are composed of a nucleus G All prokaryotes are composed of multiple cells H All prokaryotes are single celled organisms J All organisms are composed of cells
Answer:
(J) All organisms are composed of cells
At 27.0°C, the volume of a gas is 630 L. At the same pressure, its volume is 92,0 mL at a temperature of
Answer:
–272.96 °C
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 27.0 °C
Initial volume (V₁) = 630 L.
Final volume (V₂) = 92.0 mL
Final temperature (T₂) =?
Next, we shall convert 27.0 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 27.0 °C
Initial temperature (T₁) = 27.0 °C + 273
Initial temperature (T₁) = 300 K
Next, we shall convert 92.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
92 mL = 92 mL × 1 L / 1000 mL
92 mL = 0.092 L
Next, we shall determine the final temperature.
Initial temperature (T₁) = 300 K
Initial volume (V₁) = 630 L.
Final volume (V₂) = 0.092 L
Final temperature (T₂) =?
V₁ / T₁ = V₂ / T₂
630 / 300 = 0.092 / T₂
2.1 = 0.092 / T₂
Cross multiply
2.1 × T₂ = 0.092
Divide both side by 2.1
T₂ = 0.092 / 2.1
T₂ = 0.04 K
Finally, we shall convert 0.04 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
Final temperature (T₂) = 0.04 K
Final temperature (T₂) = 0.04 – 273
Final temperature (T₂) = –272.96 °C
There are four stages to the classical demographic transition model Pre-transitional Europe was characterized by high and
fluctuating mortality and a high birth rate. The transition model began to progress into and through stage 2 in the late 18th and early
19th century. All BUT ONE contributed to the decline in mortality.
S- -1]))
A)
Enacting measures to provide clean water supplies.
B)
Public health advances including quarantine of settlements undergoing
epidemics
The development of vaccines to prevent disease and antibiotics to treat
infection.
D)
Widespread acceptance of germ theory resulting in more hygienic
practices, including hand washing and sterilizing medical equipment and
infants' bottles.
Robert Hooke observed a slice of cork under a microscope. He saw what looked like “pores” or “cells” in it. Hooke can assume that the cork is most likely - Answer F A man made material G A type of living organism H A type of mineral J A petroleum based product
The cork oak tree from which cork is extracted is native to southwest europe and northwest africa. Cork is extracted from cork oak trees without harming the tree. So cork is a type of mineral. The correct option is C.
What is a cork cell?The dead cells without having intercellular spaces are defined as the cork cells. They appear at the periphery of roots and stems when they grow older and increase in girth. They also have a chemical called suberin in their walls.
It is the suberin which makes them impervious to gases and water. The outer protective coat of a tree is called the cork. It is one of the components of bark of the tree. The tissues of bark become old and the secondary mersitem replaces them.
Cork is made up of multiple thick layers and it protects the tree from bacterial or fungal infection.
Thus the correct option is C.
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The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts
Answer:
E: the point where the acid and base have been added in proper stoichiometric amounts
Explanation:
Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.
The Equivalence point occurs before the endpoint.
Thus, option E is correct.
when rolling a number cube 500 times, how many times you expect to get a 3?
Answer:
[tex]\frac{250}{3}[/tex]
Explanation:
you can expect to get a 3 (theoretically) 1 time every 6 times you roll. A 1/6 chance.
Here's the equation:
[tex]\frac{1}{6} =\frac{x}{500}[/tex]
cross multiply (i think that's what it is called)
500=6x
divide by 6 on both sides:
x=[tex]\frac{250}{3}[/tex] or approx 83 times.
Hope this helps! Lmk if u have more questions <3
In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?
Answer:
Refraction
Explanation:
200.0g of a 3.0% NaF solution, how much distilled water do we weigh out?
197g of distilled water
194g of distilled water
140g of distilled water
170g of distilled water
Answer:
194g of distilled water.
Explanation:
Hello there!
In this case, according to the given information for this problem, it turns out possible for us to use the given mass of the solution and the percent by mass of NaF to firstly calculate the grams of this solute as shown below:
[tex]\%m=\frac{m_{solute}}{m_{solution}} *100\%\\\\m_{solute}=\frac{\%m*m_{solution}}{100\%} \\\\m_{solute}=\frac{3.0\%*200.0g}{100\%} \\\\m_{solute}=6g[/tex]
And finally, since the mass of solution is calculated by adding mass of solute and mass of solvent we obtain the mass of water (solvent) as follows:
[tex]m_w=200g-6g=194g[/tex]
Therefore, the answer is 194g of distilled water
Regards!
PLEASE HELP!! ORGANIC CHEMISTRY
A sample of a diatonic gas is loaded into an evacuated bottle at STP. The 0.25 L bottle contains 1.76 grams of the unidentified gas. Calculate the molar mass of the gas. What is the identity of the diatomic gas?
Answer:
(a) 157.7 g
(b) 7.04 g/dm³
Explanation:
(a) From the question,
According to Avogadro's Law,
1 mole of every gas at STP occupies a volume of 22.4 dm³
But mass of 1 mole of the diatomic gas = molar mass of the gas.
This Implies that,
The molar mass of the gas at STP occupies a volume of 22.4 dm³
From the question,
If,
0.25 L bottle contain 1.76 g of the gas,
Therefore,
Molar mass of the gas = (1.76×22.4)/0.25
Molar mass of the gas = 157.7 g.
(b) Density of the gas = mass/volume
D = m/v
Given: m = 1.76 g, v = 0.25 L = 0.25 dm³
Therefore,
D = 1.76/0,25
D = 7.04 g/dm³
Give the ratio that balances out the following equation: ___Cr + ___Pb(NO3)4 ---> ___ Cr(NO3)3 + ____ Pb
Answer: Ratio : 4, 3, 4, 3 are the answers.
Explanation:
PLEASE HELP I HAVE 19 MINUTES LEFT I WILL MARK BRAINLIEST
How much more acidic is a pH of 4 as compared to a pH of 6.5?
Answer:
316.227766
Explanation:
What compound(s) does pure water contain?
What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe reacts with 2.1 g of O2?
4 Fe + 3O2 —> 2Fe2O3
Answer:
Fe is limiting reactant and 3.00g of Fe2O3 will be produced
Explanation:
To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:
Moles Fe -Molar mass: 55.845g/mol-
2.1g * (1mol / 55.845g) = 0.0376 moles
Moles O2 -Molar mass: 32g/mol-
2.1g * (1mol / 32g) = 0.0656 moles
For a complete reaction of 0.0656 moles of O2 are needed:
0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe
As there are just 0.0376 moles,
Fe is limiting reactant
The mass of Fe2O3 is:
Moles:
0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3
Mass:
0.0188 moles Fe2O3 * (159.69g / mol) =
3.00g of Fe2O3 will be produced
Explain what matter is, and all of the states it can have.
Answer:
matter is anything that occupies space
states of matter : solid,liquid, gas,plasma
Answer:
matter can be anything, tables chairs, literally anything. it has volume and takes up space.
Explanation:
Solids, liquids, gases, plasmas, and Bose-Einstein condensates (BEC)
Inquiry Extension Consider a reaction that occurs between solid potassium and chlorine gas. If you start with an initial mass of 15.20 g K, and an initial mass of 2.830 g Cl2, calculate which reactant is limiting. Explain how to determine how much more of the limiting reactant would be needed to completely consume the excess reactant. Verify your explanation with an example
The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].
To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.
The balanced chemical equation for the reaction between solid potassium and chlorine gas is:
2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)
From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.
First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:
moles of K = 15.20 g / 39.10 g/mol = 0.388 mol
moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol
Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:
Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl
Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl
We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.
To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.
We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:
moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])
= 0.080 mol K
This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:
mass of K needed = 0.080 mol K x 39.10 g/mol
= 3.13 g K
Therefore, The 3.13 g of K would be needed to completely react with the remaining.
Example verification:
Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?
Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]
Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol
Moles of additional [tex]Cl_2[/tex] = 0.0070 mol
The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:
0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl
Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:
5.95 g + 1.04 g = 6.99 g
The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].
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If 3.13 mol of an ideal gas has a pressure of 2.33 atm and a volume of 72.31 L, what is the temperature of the sample in degrees Celsius?
Answer:
382.49 C degree Celsius
Explanation:
Hello,
This problem deals with understanding the ideal gas law which hopes to predict how ideal gases might behave in any given condition. I listed the formula below and we are basically just going to solve for temperature by rearranging the equation as seen on the picture (there's also other rearranged ones in case you need to solve for those).
Universal gas constant R has a value of 0.0821 L * atm/(mole * K) when working with these given units so it will be part of this equation. R value changes based on what units you have.
T = PV/nR
= (2.33) (72.31) / (3.13)(0.0821)
= 655.64 K
Question is asking temperature in celsius so we employ the formula attached below:
C = K - 273.15
= 655.64-273.15
= 382.49 degree Celsius
382.49 degree Celsius is the answer!
write half-reactions that show how H2O2 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations occurred in your testing.
Answer:
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
Explanation:
When
H2O2 acts as an oxidizing agent
H2O2 + 2e- 2H+---> 2H2O
Reducing agent
H2O2 --> O2 + 2e + 2H+
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
130 cm of a gas at 20°C exerts a pressure of
750 mm Hg. Calculate its pressure if its volume
is increased to 150 cm3 at 35 °C.
Answer: The pressure is 1137.5 mm Hg its pressure if its volume is increased to 150 [tex]cm^{3}[/tex] at 35 °C
Explanation:
Given: [tex]P_{1}[/tex] = 750 mm Hg, [tex]V_{1} = 130 cm^{3}[/tex], [tex]T_{1} = 20^{o}C[/tex]
[tex]P_{2}[/tex] = ?, [tex]V_{2} = 150 cm^{3}[/tex], [tex]T_{2} = 35^{o}C[/tex]
Formula used to calculate the new pressure is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{750 mm Hg \times 130 cm^{3}}{20^{o}C} = \frac{P_{2} \times 150 cm^{3}}{35^{o}C}\\P_{2} = 1137.5 mm Hg[/tex]
Thus, we can conclude that the pressure is 1137.5 mm Hg its pressure if its volume is increased to 150 [tex]cm^{3}[/tex] at 35 °C.
1. How many grams of water are in a cup containing
0.1 moles of H2O?
Answer:
Explanation:
fdcdf
1. How does a virus differ from a common cell?
A. It has no nucleus, cell wall, or organelles.
B. It has two nuclei and no cell wall or organelles.
C. A virus has no cell well, no nucleus, and only organelles for
movement.
D. A virus differs from a cell only in shape.
Cell membranes are selectively permeable. This means that A. only water can move freely across the cell membrane. B. any substance can move across the cell membrane, but chemical energy will always be required. C. some substances can move freely across the cell membrane, while others must be transported. D. no substances can move freely across the cell membrane.
Answer:
C. some substances can move freely across the cell membrane, while others must be transported.
Explanation:
Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.
"0.0340" mol of CH₄ are in sealed flask.
Methane (CH₄)Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.
According to the question,
Volume, V = 800 mL or, 0.800 L
Temperature, T = 22°C or, 295
Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm
As we know the relation,
The gram of moles will be will be:
→ n = [tex]\frac{PV}{RT}[/tex]
By substituting the values, we get
= [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]
= [tex]\frac{0.824}{242.077}[/tex]
= 0.0340
Thus the response above is correct.
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A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution
Answer:
Following are the solution to the given question:
Explanation:
Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.
Calculate the volume of solvent present in a 55.5%
by volume of 10.5 mL alcohol solution.
Answer:
I dont know
Explanation:
good luck
A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5 °C. Find the
mass of the water.
Answer:
1.5 × 10³ g
Explanation:
Step 1: Given and required data
Transferred heat (Q): 41,840 JInitial temperature: 22.0 °CFinal temperature: 28.5 °CSpecific heat capacity of water (c): 4.184 J/g.°CStep 2: Calculate the temperature change
ΔT = 28.5°C - 22.0 °C = 6.5 °C
Step 3: Calculate the mass (m) of water
We will use the following expression.
Q = c × m × ΔT
m = Q / c × ΔT
m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g
How are tadpoles and larvae similer
Answer: Tadpole, also called polliwog, aquatic larval stage of frogs and toads. Compared with the larvae of salamanders, tadpoles have short, oval bodies, with broad tails, small mouths, and no external gills. The internal gills are concealed by a covering known as an operculum.
Explanation:
A balloon is inflated to a volume of 8.0 L on a day when the atmospheric pressure is 1.013 bar . The next day, a storm front arrives, and the atmospheric pressure drops to 0.968 bar . Assuming the temperature remains constant, what is the new volume of the balloon, in liters
Answer:
[tex]V_2=8.4L[/tex]
Explanation:
Hello there!
In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:
[tex]baV_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{8.0L*1.013bar}{ 0.968bar}\\\\V_2=8.4L[/tex]
Regards!