Internal structures that helps the plant grow

Option (4) is correct. The rate constant of a second order reaction has the unit  L/mole. sec.

In the Second order reaction the rate is proportional to the square of the concentration of one reactant. Rate of Second order reaction  is proportional to the product of the concentrations of two reactants. Such reactions generally have the form,

A + B → products.

Each monomer combines to form a larger molecule is called dimer. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mole/L, the unit of the rate constant can also be written as L(mole ·s).

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To solve this issue, we must use stoichiometry to calculate the relationship between the production of CO2 and the amount of C3H8.

Using its molar mass, we must first determine the moles of C3H8: C3H8 has a molar mass of 44.11 g/mol, which is calculated by multiplying 3 (12.01 g/mol) by 8 (1.01 g/mol). C3H8 moles are equal to 11.89 g / 44.11 g/mol, or 0.27 mol. Next, we apply the balanced chemical equation's C3H8 to CO2 mole ratio: Three molecules of CO2 are produced from one mol of C3H8. the following will be the moles of CO2 produced: The formula for moles of CO2 is: 0.27 mol C3H8 x (3 mol CO2/1 mol C3H8) = 0.81 mol CO2. Finally, we use its molar mass to translate the moles of CO2 into grammes: Molar mass of CO2 equals 12.01 g/mol plus 2 (16.00 g/mol), or 44.01 g/mol. mass of CO2 = 44.01 g/mol x 0.81 mol CO2 to get 35.64 g CO2. As a result, 11.89 g of C3H8 will result in 35.64 g of CO2.

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Answer:

Height of tray

Explanation:

The answer is B 12.5g

Out of the given alkyl halides, the fourth option, CH3N(CH3)CH2CH2Br, could be successfully used to form a Grignard reagent. This is because Grignard reagents are formed by the reaction of an alkyl halide with magnesium metal in dry ether.

The resulting product is an organomagnesium compound, which can be further used in organic synthesis. In option 1, the hydroxyl group would interfere with the reaction, while in option 2, the amino group would also hinder the reaction. Option 3 has a carboxylic acid group, which is not an alkyl halide. Therefore, option 4 is the only suitable alkyl halide for forming a Grignard reagent.

A Grignard reagent is formed by reacting an alkyl halide with magnesium in an ether solvent. To be successful, the alkyl halide must not contain any acidic protons (H) that can react with the Grignard reagent. In the given options, 1. OHCH2CH2CH2CH2Br and 3. BrCH2CH2CH2COOH have acidic protons in their alcohol and carboxylic acid groups, respectively, and thus cannot form a Grignard reagent. 2. H2NCH2CH2CH2Br has an acidic proton in the amine group and is also unsuitable. The only suitable option is 4. CH3N(CH3)CH2CH2Br, as it lacks any acidic protons and can successfully form a Grignard reagent.

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Answer:

Radiant energy passes across Earth's system boundary, but not mass.

Explanation:

In a closed system, no substance can escape or be added to it. Energy, on the other hand, can still reach the system or leave the system again.

Answer: radiant energy passes across Earth's system boundary, but not mass

Explanation:

100% K12 test

The systematic name for the compound Ba(NO3)2 is barium nitrate. Barium nitrate is an inorganic salt with the chemical formula Ba (NO3)2. It is a colorless, odorless, and crystalline solid that is highly soluble in water. The compound is formed by combining one atom of barium and two ions of nitrate.

The name “barium” comes from the Greek word “barys,” which means “heavy,” and is a reference to its high density. The term “nitrate” refers to the polyatomic ion NO3-, which is composed of one nitrogen atom and three oxygen atoms. Barium nitrate is commonly used in pyrotechnics, as it is a powerful oxidizing agent that produces a bright green flame when ignited.

The systematic naming of inorganic compounds is based on the rules set out by the International Union of Pure and Applied Chemistry (IUPAC). The name of an ionic compound is composed of the cation name followed by the anion name. In the case of barium nitrate, “barium” is the name of the cation, while “nitrate” is the name of the anion.

Therefore, the systematic name for the compound Ba(NO3)2 is barium nitrate.

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Answer:

B) mostly empty space and has a small, positively charged nucleus

Explanation:

In the gold foil experiment, positively-charged alpha particles were directed towards a gold foil sheet. During the experiment, most of the particles went through the gold foil. However, a select few alpha particles were met with resistance and bounced off the sheet.

This proves that the gold atoms, which made up the gold foil sheet, were mostly empty space as most of the alpha particles passed through it. Furthermore, the particles which bounced off the sheet must have hit small, positively-charged nuclei. The nuclei must have been positive because similar charges repel each other. In other words, if the nuclei were negatively-charged, the positively-charged alpha particles would not bounce off the sheet, but instead "stick" to it.

From the collected three samples from the coral reef we can conclude that:

SAMPLE A - pure substance.

SAMPLE B - homogeneous mixture.

SAMPLE C - heterogeneous mixture.

Pure substances and mixtures are the two broad categories into which matter can be divided.

A sort of matter known as a pure substance has qualities that are constant throughout the sample and a stable composition that makes it the same everywhere (meaning that there is only one set of properties such as melting point, color, boiling point, etc. throughout the matter).

A mixture is said to be homogenous if its composition is constant throughout. Because the dissolved salt is evenly distributed throughout the whole salt water sample, the salt water described above is homogenous.

A combination is said to be heterogeneous if its composition is not constant throughout. Vegetable soup is a complex concoction. Each mouthful of soup will have differing percentages of the various veggies and other ingredients.

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The concentration of the solution is 11.79 mol/L (or 11.79 M) of MgCl2.

The following steps should be followed in general to determine a compound's molar mass:

Indicate the number of each type of atom in the molecule when writing the compound's chemical formula.

Get the atomic mass of each element by consulting the periodic table.

The atomic mass of each element is multiplied by the quantity of that element's atoms in the molecule.

Add up the products from step 3 to get the total molar mass of the compound.

To determine the concentration of a solution, we need to know the amount of solute (in grams) and the volume of the solution (in liters).

First, we need to convert the volume of the solution from milliliters (mL) to liters (L):

18 mL = 18/1000 L = 0.018 L

Next, we need to calculate the concentration of MgCl2 in the solution:

concentration = amount of solute / volume of solution

In this case, the amount of solute is 20.2 grams of MgCl2. We can use the molar mass of MgCl2 to convert from grams to moles:

molar mass of MgCl2 = 95.21 g/mol

moles of MgCl2 = 20.2 g / 95.21 g/mol = 0.2122 mol

Now we can calculate the concentration:

concentration = 0.2122 mol / 0.018 L

concentration = 11.79 mol/L

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The p-value is greater than the significance level of 0.05, we do not reject the null hypothesis and conclude that all proportions are equal.

Firstly, let us conduct a Chi-square test of independence of categorical variables based on the information given above. We have three different cases of hypothesis testing that we have to solve one by one.

Case 1: HD​:pSun ​=rhoMan ​=pTue ​=pWed ​=pThu ​=pFri ​=pSat ​=71​
Ha​ : Not all proportions are equal.

Test Statistic
For this hypothesis, we need to compute the test statistic that is given as:
\($$\chi^2=\sum_{i=1}^{k}\frac{(O_i-E_i)^2}{E_i}$$\)  where k is the number of groups/categories. Since we have 7 days of the week, \(k = 7. $O_i$ and $E_i$\) are the observed and expected frequencies respectively.  
Here, we have equal proportions of 71 for each day of the week.
Therefore, the expected frequencies are also equal to 71.
\($$E_i = 71, i=1,2,3,4,5,6,7.$$\)

We also have to use the given information to compute the observed frequencies,
\($O_i$.$$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126$$\)

Therefore, the test statistic can be computed as \($$\chi^2=\frac{(90-71)^2}{71} + \frac{(99-71)^2}{71} + \frac{(122-71)^2}{71} + \frac{(123-71)^2}{71} + \frac{(130-71)^2}{71} + \frac{(160-71)^2}{71} + \frac{(126-71)^2}{71}$$$$\chi^2=180.14\)

Now we have to find the p-value of this test. Since the number of degrees of freedom is k - 1 = 7 - 1 = 6, the p-value can be found using the chi-square distribution table with 6 degrees of freedom at the 0.05 significance level. The p-value is 0.000014. ConclusionSince the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that not all proportions are equal.

The total number of accidents is \($$90+99+122+123+130+160+126=850$$\)

The percentage of accidents occurring on each day of the week can be found as follows:
\($$Sunday: $$\frac{90}{850}\times 100 = 10.59\%$$Monday: $$\frac{99}{850}\times 100 = 11.65\%$$Tuesday: $$\frac{122}{850}\times 100 = 14.35\%$$Wednesday: $$\frac{123}{850}\times 100 = 14.47\%$$Thursday: $$\frac{130}{850}\times 100 = 15.29\%$$Friday: $$\frac{160}{850}\times 100 = 18.82\%$$Saturday: $$\frac{126}{850}\times 100 = 14.82\%$$\)

From the above percentages, we can see that Friday has the highest percentage of traffic accidents.

Case 2:
HD​: Not all proportions are equal.

Ha​:pSun ​=pMon ​=pTue ​=rhoWed ​=pThu ​=rhoFri ​=rhoSat ​=71

Test Statistic

\($$E_1 = 78.57, E_2 = 86.57, E_3 = 106.86, E_4 = 107.43, E_5 = 113.57, E_6 = 139.43, E_7 = 109.14$$\)

We already know the observed frequencies,
\($$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126.$$\)

The test statistic can be computed as:

\($$\chi^2=\frac{(90-78.57)^2}{78.57} + \frac{(99-86.57)^2}{86.57} + \frac{(122-106.86)^2}{106.86}+ \frac{(123-107.43)^2}{107.43} + \frac{(130-113.57)^2}{113.57} + \frac{(160-139.43)^2}{139.43} + \frac{(126-109.14)^2}{109.14} $$$$ \implies \chi^2=34.98$$\)
The p-value is 0.000001.
Conclusion- Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that not all proportions are equal.

Case 3:

All proportions are equal.
Test Statistic
The expected frequency for each group is
\(E = \frac{850}{7} = 121.43\)
We already know the observed frequencies,
\($$O_1 = 90, O_2 = 99, O_3 = 122, O_4 = 123, O_5 = 130, O_6 = 160, O_7 = 126.$$\)

The test statistic is,

\($$\chi^2=\frac{(90-121.43)^2}{121.43} + \frac{(99-121.43)^2}{121.43} + \frac{(122-121.43)^2}{121.43} + \frac{(123-121.43)^2}{121.43} + \frac{(130-121.43)^2}{121.43} + \frac{(160-121.43)^2}{121.43} + \frac{(126-121.43)^2}{121.43}} \\\implies \chi^2=9.17$$\)

The p-value is 0.1664.

Since the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis and conclude that all proportions are equal.

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Answer:

the kidney control the amount of water, ion, and other substances in the blood by excreting more or less of them in urine, also secretes the hormones that help maintain homeostasis

Answer:

ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation. Example 4.8. Moles of Reactant Required in a Reaction.

Explanation:

First reaction: Bi-211 (f) → Tl-207 (a) + α (i), where α is an alpha particle. Second reaction: Tl-207 (w) → Pb-207 (stable isotope) + β (z), where β is a beta particle.


In the first reaction, bismuth-211 (Bi-211) decays through alpha emission, producing thallium-207 (Tl-207) and an alpha particle (α). The equation is:
Bi-211 (f) → Tl-207 (a) + α (i)

In the second reaction, thallium-207 (Tl-207), which was produced in the first reaction, decays through beta emission to form a stable isotope, lead-207 (Pb-207), and a beta particle (β). The equation is:
Tl-207 (w) → Pb-207 (stable isotope) + β (z)
These two equations represent the nuclear reactions that occur as bismuth-211 decays to a stable isotope through both alpha and beta emissions.

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H2po4-1 has a pka of 6.86. if the concentration of h2po4-1 is 0.622 mm and the concentration of h1po4-2 is 0.038 mm the ph value in the solution is 8.3

pH stand for potential hydrogen the pH of a solution also can be determined by finding the pOH. determine the concentration of the hydroxide ions by dividing the molecules of hydroxide by the volume of the solution and here to find pH value in the solution here the formula is

pH=pka log acid/base

ph=pka log  h2po4-1/h1po4-2

ph=6.86×log0.622/ 0.038

ph=6.86×log16.36

ph=6.86×1.21

ph=8.3

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Answer:

\(\boxed {\boxed {\sf 9.6 \ mol \ KCl}}\)

Explanation:

We must use stoichiometry to solve this, which is the calculation of reactants and products in a reaction using ratios.

Let's analyze the reaction given.

\(MgCl_2 _{(aq)} + K_2SO_4 _{(aq)} \rightarrow 2KCl _{(aq)} + MgSO_4 _{(s)}\)

Now, look at the coefficients, or numbers in front of the molecule formulas. If there isn't a coefficient, then a 1 is implied.

We want to find how many moles of potassium chloride (KCl) are produced from 4.8 moles of magnesium chloride (MgCl₂). Check the coefficients for these molecules.

The coefficient represents the number of moles. Therefore, 1 mole of magnesium chloride produces 2 moles of potassium chloride. We can set up a ratio using this information.

\(\frac { 1 \ mol \ MgCl_2} {2 \ mol \ KCl}\)

Multiply by the given number of moles of magnesium chloride: 4.8

\(4.8 \ mol \ MgCl_2 *\frac { 1 \ mol \ MgCl_2} {2 \ mol \ KCl}\)

Flip the ratio so the moles of magnesium chloride cancel out.

\(4.8 \ mol \ MgCl_2 *\frac {2 \ mol \ KCl} { 1 \ mol \ MgCl_2}\)

\(4.8 *\frac {2 \ mol \ KCl} { 1 \ } }\)

\(4.8 * {2 \ mol \ KCl}\)

\(9.6 \ mol \ KCl\)

9.6 moles of potassium chloride are produced from 4.8 moles of magnesium chloride.

Answer:

B. a ball in someones hands

Explanation:

Because this ball possess energy due to virtue of its position.

Answer:

B

Explanation:

It can be moved at any point in somebody's hands so it will be B

The percent abundance of the ag-107 isotope if the element silver (atomic mass 107.868 amu) has two naturally occurring isotopes, ag-107 (mass 106.9051 amu) and ag-109 (mass 108.9048 amu) is 51.8%.

To determine the percent abundance of the ag-107 isotope, we can find it by using the formula:

percent abundance = (number of atoms of isotope ÷ total number of atoms of all isotopes) × 100

Firstly, we will calculate the average atomic mass of silver as it is the average of masses of its isotopes. Now, we have to find the percentage of each isotope and then add them up. So, let's calculate them:

The average atomic mass of silver is Average atomic mass of silver = [(mass of isotope 1) × (% abundance of isotope 1) + (mass of isotope 2) × (% abundance of isotope 2)] / 100107.868 amu

= (106.9051 amu × x) + (108.9048 amu × (1 – x))107.868 amu

= 106.9051x + 108.9048 – 108.9048x

x = 0.518 (rounded to 3 decimal places) and 1 – x = 0.482

So, percent abundance of ag-107 isotope = 51.8%.

Thus, the percent abundance of the ag-107 isotope is 51.8% (0.518 in decimal notation) rounded to 1 decimal digit.

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The intermolecular forces between molecules in a liquid sample of sulfur trioxide, SO3 is London dispersion force.

The electrostatic forces of attraction present between molecules which hold them together are termed as intermolecular force.

This type of forces arises between symmetrical non- polar molecules. At a particular instant of time, the symmetry of molecules break and become unsymmetric. One side contain more electron than other. Having more electron become negative and other side get negative.

This unsymmetry make neighbouring molecule unsymmetric. The weak dispersion bond is formed between molecules called london dispersion force.

Thus, intermolecular forces between molecules in a liquid sample of sulfur trioxide, SO3 is london dispersion force.

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We need 0.121 mL of 0.108897 M NaOH solution to titrate 0.227 g of K\(KHC_{8}H_{4}O_{4}\)

The balanced equation for the reaction between sodium hydroxide (NaOH) and potassium hydrogen phthalate (KHC8H4O4) is:

NaOH + \(KHC_{8}H_{4}O_{4}\) → \(NaKC_{8}H_{4}O_{4}\) + H2O

Number of moles of \(KHC_{8}H_{4}O_{4}\) = mass ÷ molar mass

= 0.227 g ÷ 204.22 g/mol

= 0.001111 mol

Volume of NaOH = Number of moles of NaOH × Molarity of NaOH

We can calculate the number of moles of NaOH required to react with 0.001111 mol of \(KHC_{8}H_{4}O_{4}\)as follows:

Number of moles of NaOH = Number of moles of KHC8H4O4

= 0.001111 mol

Substituting the values into the equation, we get:Volume of NaOH = 0.001111 mol × 0.108897 M

= 0.000121 L or 0.121 mL (since 1 L = 1000 mL)

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Diatoms are single-celled algae that have a silica (silicate) shell called a frustule.

Diatoms are photosynthetic organisms and are known for their intricate and diverse shapes. Diatoms are commonly found in freshwater and marine environments and play a significant role in the global carbon cycle.

Micrite is a fine-grained carbonate sedimentary rock composed of tiny carbonate particles. It forms through the precipitation and accumulation of carbonate minerals, such as calcite or aragonite, in marine environments. In the case of chalk, which is primarily composed of microscopic fragments of calcium carbonate from marine organisms, recrystallization can occur at the ocean bottom under specific conditions, leading to the formation of micrites.

Therefore, it's important to note that chert, marble, and quartzite are not the typical products of recrystallization of chalk at the ocean bottom.

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Answer:

B. False

Explanation:

step by step explanation

Answer:

False

Explanation:

Answer:

ΔG°rxn = 28.4kJ/mol at 600K

Explanation:

Using Hess's law, you can find the ΔH°rxn and S°  subtracting ΔH°f of products - ΔH°f of reactants ×its coefficients. In the same way for S°rxn

For example, for the reaction:

aA + bB → cC:

ΔH°rxn = c×ΔH°fC - (a×ΔH°fA + b×ΔH°fB).

S°rxn = c×S°fC - (a×S°fA + b×S°fB).

For the reaction:

SO₂(g) + Cl₂(g) → SO₂Cl₂(g)

ΔH°rxn = 1×ΔH°f{SO₂Cl₂} - (1×ΔH°fSO₂ + 1×ΔH°fCl₂).

S°rxn = 1×S°f{SO₂Cl₂} - (1×S°fSO₂ + 1×S°fCl₂).

As at 298K:

ΔH°f{SO₂Cl₂} = -364.0kJ/mol

ΔH°f{SO₂} = -296.8kJ/mol

ΔH°f{Cl₂} = 0kJ/mol

ΔH°rxn = 1×{-364.4kJ/mol} - (1×-296.8kJ/mol + 1×0).

ΔH°rxn = -67.2kJ/mol at 298K.

S°f{SO₂Cl₂} = 311.9J/molK

S°f{SO₂} = 248.2J/molK

S°f{Cl₂} = 223.0J/molK

S°rxn = 1×{311.9J/molK} - (1×248.2J/molK + 1×223.0J/molK).

S°rxn = -159.3J/molK = -0.159.3KJ/molK

Using:

ΔG°rxn = ΔH°rxn - S°rxn×T

Assuming ΔH°rxn doesn't change at 600K:

ΔG°rxn = -67.2kJ/mol - -0.159.3J/molK×600K

Sodium from Na3PO4(aq) could be prepared by electrolysis of the aqueous solution shown. Based on the provided options, the element that could be prepared by electrolysis of the aqueous solution shown

Potassium from KCl(aq)
Here's why:
- Sodium from Na3PO4(aq) and Nitrogen from AgNO3(aq) are not possible because these ions are more stable in solution than undergoing electrolysis.
- Sulfur from K2S04(ed) is not valid as the compound should be K2SO4(aq) and even then, it would produce oxygen at the anode instead of sulfur.
- Oxygen from H2SO4(aq) can be prepared through electrolysis, but this is not an element directly obtained from the compound.
Potassium from KCl(aq) can be prepared by electrolysis. During this process, K+ ions are reduced to potassium metal at the cathode, and Cl- ions are oxidized to chlorine gas at the anode.

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Answer:

Intensive, it does not depend on how much matter is present

so like.. no mater how much water you are freezing.. the freezing point will always be the same

Answer:

An alkaline solution of glucose acts as a reducing agent and reduces added methylene blue from a blue to a colourless form.