How many protons and neutrons are in an atom of P-32

Na, C, and O so the answer is c. 3

The molar mass of aluminum sulfate is 342.15 g/mol.

To find the number of moles in the solution, we first need to calculate the mass of aluminum sulfate present in 1 liter of the solution:

0.284 g/L x 1 L = 0.284 g

Then we can find the number of moles:

0.284 g / 342.15 g/mol = 0.000829 moles

Since we only have 396 mL of the solution, we need to adjust the number of moles accordingly:

0.000829 moles/L x 0.396 L = 0.000328 moles

Aluminum sulfate dissociates in water to form 3 ions: one aluminum ion (Al3+) and two sulfate ions (SO42-).

So for every 1 mole of aluminum sulfate dissolved, we get 1 mole of Al3+ ions and 2 moles of SO42- ions.

Therefore, in 0.000328 moles of aluminum sulfate, we have:

0.000328 moles of Al3+ ions

0.000656 moles of SO42- ions

To find the number of ions of each type, we can use Avogadro's number, which tells us the number of particles (ions, atoms, molecules) in 1 mole of a substance:

Number of Al3+ ions: 0.000328 moles x 6.022 x 10^23 ions/mol = 1.97 x 10^20 ions

Number of SO42- ions: 0.000656 moles x 6.022 x 10^23 ions/mol = 3.95 x 10^20 ions

Therefore, there are approximately 1.97 x 10^20 aluminum ions and 3.95 x 10^20 sulfate ions in 396 mL of 0.284 g/L aluminum sulfate solution.

(a) Van der Waal's forces. Van der Waal's forces are weak electrostatic interactions between non-polar molecules.

These weak forces arise from the fluctuating dipoles within the molecules, which cause temporary charges to develop and attract each other. This attraction leads to the formation of weak intermolecular bonds, which are mainly responsible for the attraction between non-polar molecules. These forces are weak in comparison to other intermolecular forces, such as hydrogen bonding and ionic bonds, but are still important for the stability of non-polar molecules and their ability to dissolve in other non-polar solvents.Van der Waals forces are attractive intermolecular forces between molecules caused by the fluctuating dipole moments of molecules.

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complete question:The attraction that non-polar molecules have for each other is primarily caused by —

(a) Van der Waal's forces (b) Difference in electronegativities

(c) Hydrogen bonding (d) High ionisation energy​

Answer:

12

Explanation:

12

To solve this we must be knowing each and every concept related to hybridisation. Therefore, the hybridisation of the given molecule can be given as below.

Only during bond formation does hybridization occur, not in an individual gaseous atom. If the bonding of a molecule is known, the geometry of the molecule may be predicted.

compound                                                              hybridisation

Carbon dioxide (0 = C = O)                                       sp

Propene (CH\(_3\)CH = CH\(_2\))                                            sp³, sp²,sp²

Formaldehyde (H\(_2\)C = O)                                             sp²

Acetone [(CH\(_3\) )\(_2\) C = O                                                 sp³, sp²,      sp³

Ketene (H\(_2\)C = C = O)                                                  sp²,    sp

Therefore, the hybridisation of the given molecule can be given as above.

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3 Cu + 8HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O

n(HNO3)=6.3mol

unknown/known=4/8=1/2

n(H2O)=n(HNO3)/2=3.15mol

m(H2O)=nxM=3.15x(1+1+16)=56.7g

Let me know if its wrong

Answer:

I’m getting 1 hour and 50 minutes

Explanation:

Answer:

its Teaching, research, design.

Explanation:

hope this helps :)

Answer: 50mL

Explanation: This is solved just like you'd solve a proportion in mathematics. But first, you need to make your temperature values have the same units. It is ideal to have both units in K almost 90% of the time. So to convert the values, you'll add 273 to your Celsius temperature: 27. This gives you 300. Then, you set 300K/25mL = 600K/?. We see that to get from 300 to 600, we just double our values. Therefore, we'll do the same to the 25mL. This gives us 50mL.

I hope this helps!

Answer:

Hydrogen fuel typically reacts with oxygen in a chemical reaction to produce water (H2O) as the main product. The chemical equation for this reaction is:

2H2 + O2 → 2H2O

The reaction releases energy, which can be harnessed to power engines or generate electricity. When hydrogen fuel is burned in an engine or fuel cell, the only byproduct is water vapor, which is harmless to the atmosphere.

However, the production of hydrogen fuel can have environmental impacts if it is not produced using renewable energy sources. The most common method of producing hydrogen involves using natural gas, which releases carbon dioxide and other greenhouse gases into the atmosphere. Therefore, it is important to consider the source of the hydrogen fuel and the environmental impact of its production when evaluating its benefits for reducing emissions and improving air quality.

The chemical equation for the combustion of hydrogen can be represented as follows: 2H₂ + O₂ → 2H₂O

In this reaction, two molecules of hydrogen gas (H₂) combine with one molecule of oxygen gas (O₂) to form two molecules of water vapor (H₂O). The reaction releases a significant amount of energy in the form of heat.

The products of the combustion of hydrogen, which are water molecules, do not have significant adverse effects on the atmosphere. Water vapor is a natural component of the Earth's atmosphere, and its presence is essential for various natural processes, such as the water cycle.

However, it is important to note that the combustion of hydrogen as a fuel can indirectly affect the atmosphere in certain scenarios. One consideration is the generation of the primary energy used to produce the hydrogen fuel.

Additionally, the transportation and storage of hydrogen as a fuel may present challenges. If there are any leaks or improper handling, hydrogen gas can contribute to the formation of local air pollution. However, these issues can be managed through appropriate safety measures and technology advancements.

Overall, the combustion of hydrogen fuel with oxygen produces water as a product, and the water vapor does not have direct adverse effects on the atmosphere.

However, the environmental impact associated with hydrogen as a fuel depends on the methods used for its production and transportation. Sustainable production methods, such as electrolysis using renewable energy sources, can minimize the environmental impact of hydrogen fuel.

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Answer:

The problem provides you with the solubility of potassium chloride,

KCl

, in water at

20

∘

C

, which is said to be equal to

34 g / 100 g H

2

O

.

This means that at

20

∘

C

, a saturated solution of potassium chloride will contain

34 g

of dissolved salt for every

100 g

of water.

As you know, a saturated solution is a solution that holds the maximum amount of dissolved salt. Adding more solid to a saturated solution will cause the solid to remain undissolved.

In your case, you can create a saturated solution of potassium chloride by dissolving

34 g

of salt in

100 g

of water at

20

∘

C

.

Now, your goal here is to figure out how much potassium chloride can be dissolved in

300 g

of water at this temperature. To do that, use the given solubility as a conversion factor to take you from grams of salt to grams of water

Answer:

7.51

×

10

24

The density

ρ

of a sample is its ratio of mass

m

to volume

V

:

ρ

=

m

V

.

To find the number of atoms in this sample, first multiply the volume by the density to get the mass. Then divide this mass by the mass of one atom.

133

c

m

3

T

a

(

17.0

g

1

c

m

3

)

(

1

atom

T

a

3.01

×

10

−

22

g

T

a

)

=

7.51

×

10

24

Ta atoms

To determine the specific heat of the unknown substance, we can use the principle of energy conservation. The heat lost by the unknown substance is equal to the heat gained by the water.Therefore, the specific heat of the unknown substance is approximately 2.68 J/g°C.

First, we calculate the heat lost by the unknown substance. Using the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature, we have:

Q_substance = m_substance * c_substance * ΔT_substance

Next, we calculate the heat gained by the water. Again, using the same formula, we have:

Q_water = m_water * c_water * ΔT_water

Since the system reaches equilibrium, Q_substance = Q_water. Rearranging the equation and substituting the given values, we can solve for c_substance:

m_substance * c_substance * ΔT_substance = m_water * c_water * ΔT_water

c_substance = (m_water * c_water * ΔT_water) / (m_substance * ΔT_substance)

Plugging in the values, we find:

c_substance = (250.00 g * 4.18 J/g°C * (29.000 °C - 23.000 °C)) / (95.00 g * (98.000 °C - 23.000 °C))

c_substance ≈ 2.68 J/g°C

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Answer:

Acetic acid, carbonic acid and chlorous acid with calcium hydroxide

Ammonia with sulfuric acid

Explanation:

A buffer is an aqueous mixture of a weak acid and its conjugate base or vice versa.

Weak acids reacts with strong bases to produce the conjugate base. In the right amount, you can produce a buffer. In the same way, you can produce a buffer from the mixture of weak bases with strong acids.

In the problem, you have weak acids (acetic acid, carbonic acid, chlorous acid), one weak base (ammonia), one strong base (calcium hydroxide) and one strong acid (Sulfuric acid).

Thus, the mixtures that can produce a buffer are:

Acetic acid, carbonic acid and chlorous acid with calcium hydroxide

And:

Ammonia with sulfuric acid

The pH at the equivalence point for the titration of 0.20 M HCl versus 0.10 M NaOH is 7.0.

The reaction between HCl and NaOH is a strong acid-strong base titration. At the equivalence point, all of the HCl has reacted with an equal amount of NaOH to form water and NaCl.

The balanced equation for the reaction is:

HCl + NaOH → NaCl + H2O

Since the reaction between HCl and NaOH is a 1:1 stoichiometric ratio, we can say that at the equivalence point, the moles of HCl consumed are equal to the moles of NaOH added.

Moles of HCl = concentration of HCl × volume of HCl used

Moles of NaOH = concentration of NaOH × volume of NaOH added

At the equivalence point, Moles of HCl = Moles of NaOH. Therefore,

concentration of HCl × volume of HCl used = concentration of NaOH × volume of NaOH added

Since we have equal volumes of HCl and NaOH at the equivalence point, we can simplify the equation to:

concentration of HCl = concentration of NaOH

Therefore, the concentration of HCl and NaOH at the equivalence point is both 0.15 M (the average of the initial concentrations of 0.20 M HCl and 0.10 M NaOH).

To find the pH at the equivalence point, we can use the equation for the dissociation of water:

H2O ⇌ H+ + OH-

At 25°C, the concentration of water is 55.5 M. At the equivalence point, the concentration of H+ and OH- are both equal, so we can write:

Kw = [H+][OH-]

where Kw is the ion product constant for water, which has a value of 1.0 × 10^-14 at 25°C.

At the equivalence point, the concentration of H+ and OH- are both equal to the concentration of NaOH and HCl, which is 0.15 M.

Therefore:

Kw = [H+][OH-]

1.0 × 10^-14 = [0.15][0.15]

[H+] = [OH-] = sqrt(1.0 × 10^-14) = 1.0 × 10^-7 M

The pH at the equivalence point is equal to the negative logarithm of the H+ concentration:

pH = -log[H+]

pH = -log(1.0 × 10^-7)

pH = 7.0

Therefore, the pH at the equivalence point for the titration of 0.20 M HCl versus 0.10 M NaOH is 7.0.

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Answer:

star a is farther away

Explanation:

Answer:

the mass numder of atoms is called atomic mass

Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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A) Calculate the lower and upper bounds of the 95% confidence interval:

Lower bound = X - ME

Upper bound = X + ME

B) you would compare the obtained mean concentration (0.101675 N) with the prepared concentration (0.1012 N) and calculate the t-value. The t-value is then compared to the critical t-value at a certain significance level (e.g., a = 0.05) and the degrees of freedom (n-1) to determine if there is a statistically significant difference.

(a) To determine the range within which you can be 95% confident that the true value falls, you can calculate the confidence interval using the obtained results.

First, calculate the mean (X) of the measured concentrations:

X = (0.1017 + 0.1019 + 0.1016 + 0.1015) / 4 = 0.101675 N

Next, calculate the standard deviation (s) of the measured concentrations:

s = sqrt(((0.1017 - X)² + (0.1019 - X)² + (0.1016 - X)² + (0.1015 - X)²) / (4 - 1))

Then, calculate the standard error of the mean (SE) using the formula:

SE = s / sqrt(n)

where n is the number of measurements (in this case, n = 4).

With the standard error, you can calculate the margin of error (ME) for a 95% confidence interval using the t-distribution. The t-value for a 95% confidence interval with 3 degrees of freedom is approximately 3.182.

ME = t * SE

Finally, calculate the lower and upper bounds of the 95% confidence interval:

Lower bound = X - ME

Upper bound = X + ME

(b) To determine if there is a statistical difference between the obtained mean concentration (0.101675 N) and the prepared concentration (0.1012 N), you can perform a t-test. The t-test compares the means of two sets of data to determine if they are significantly different from each other.

Using the formula for a two-sample t-test, you would compare the obtained mean concentration (0.101675 N) with the prepared concentration (0.1012 N) and calculate the t-value. The t-value is then compared to the critical t-value at a certain significance level (e.g., α = 0.05) and the degrees of freedom (n-1) to determine if there is a statistically significant difference.

If the calculated t-value is greater than the critical t-value, it suggests a statistically significant difference between the obtained mean concentration and the prepared concentration.

Please note that the actual calculations and interpretation may vary depending on the statistical software or method used, and it's always recommended to consult a statistician or refer to appropriate statistical references for accurate analysis.

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1. We are given :

• n = 4.12 mol

• at STP , we know that :

According to avogadro's law at STP 1 mole of gas occupy volume 22.414 L

2. Calculations :

If 1 mole : 22.424L

then,4.12mole : x L .......(do a cross multiply)

Therefore ; x L = 4.12Mole * 22.424L/ 1 mole

=92.38 Litres

≈92.4 L

• This means that 4.12 mol sample of hydrogen gas will occupy ,92.4 Litres

try stopping putting the car in park before fixing

Answer:

4-5

Explanation:

maybe 5

The revenues of the note will equal the $225,000 face value, provided the note has an interest rate of 8%.

On Friday, December 9, 2022, the benchmark 30-year fixed mortgage's average rate is 7.32%, which is a 15 basis point increase from the previous week. If you want to restructure your present mortgage.

The nominal value and face value are both used interchangeably. That which a money is worth. Simply said, it is a security's nominal value. A 45-day note having a face value of $225,000 was issued by Nyree Co. to Conrad Bank based on the information provided. Consequently, the amount of the note's revenues will match its face value, which is 225,000.

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The usual enthalpy change for the 200g/7.68 mol of ethyne reaction is C2H6g C2H2g Plus 2H2g - 2,388.5kJ.

The quantity of heat released or absorbed during a reaction occurring at constant pressure is known as the enthalpy change.

H stands for "enthalpy change," Hf for "system final enthalpy" (i.e., the enthalpy of the byproducts of the system in equilibrium in a chemical reaction), and Hi for "system initial enthalpy" (i.e., the entropy for the reactants in a chemical reaction).

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Answer:

Physical Property

Explanation:

Density, mass, volume, color, melting and boiling points, etc. are all physical properties. No matter what changes, the chemical makeup stays the same.

Flamability, acidity, toxicity, etc. are chemical properties, because they chemically change the makeup of the object/thing.

15.6 grams of water are formed when 500.0 kJ of energy is released in the synthesis of water from its elements.

The balanced chemical equation for the synthesis of water from its elements is:

O2(g) + 2H2(g) → 2H₂O(l) + 572 kJ

The given energy change is -500.0 kJ, which means that energy is released during the reaction.

We can use the relationship between the amount of energy released and the amount of substance involved in the reaction to calculate the mass of water formed.

The molar enthalpy change for the reaction can be calculated as follows:

ΔH = -500.0 kJ/mol H2O

The molar enthalpy change tells us how much energy is released when 1 mole of water is formed.

The molar mass of water (H2O) is 18.015 g/mol.

To calculate the mass of water formed, we can use the following equation:

mass of water = (ΔH / molar enthalpy change) x molar mass of water

mass of water = (-500.0 kJ / mol H2O / -572 kJ/mol) x 18.015 g/mol

mass of water = 15.6 g

Therefore, 15.6 grams of water are formed when 500.0 kJ of energy is released in the synthesis of water from its elements.

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