pls help asap !! unsure how to do this

Answer:

C

Explanation:

this is because the rest talk about the findings of protons,neutrons and electrons.

Answer:

Liquid

Explanation:

We can rule out solids when there is no definite shape, so we have liquid and gas left. The thing is gas doesn't have a definite volume because we can't transfer it between two glass cups (this is an example). Thus, liquid is the state of matter in which a material has a definite volume but not a definite shape.

Number of particles : 8.603 x 10²³

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

Assumption⇒ Standard Conditions

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

1 mol = 6.02 x 10²³ particles (atoms, ions, or molecules)

32.0 liters of Helium

\(\tt mol=32:22.4=1.429\)

Number of particles :

\(\tt 1.429\times 6.02\times 10^{23}\\\\=8.603\times 10^{23}\)

When a chemical is spilled on the skin or eyes, you should immediately flood the area with plenty of running water. This is the first and most important step in treating chemical spills on the skin or eyes.

Flushing the affected area with running water helps to dilute and remove the chemical from the skin or eyes, effectively minimizing its contact and potential harm. It is crucial to continue rinsing the area for at least 15 to 20 minutes to ensure thorough flushing.

This prompt action of flooding the area with running water helps to reduce the potential damage caused by the chemical and promotes the removal of any remaining chemical residues. It is important to seek medical attention immediately after flushing, especially if the chemical is hazardous or the symptoms persist.

Remember, in case of a chemical spill on the skin or eyes, the priority is to flush the affected area with plenty of running water without delay.

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An example of observation that would provide a scientist with the best data for planning a new investigation about plant growth is; Choice C; The scientist observed that after about a month, the green plant in some soil looked taller than the rest of the plants.

Discussion:

An observation like that in Choice C where in; what soil the green plant looked taller than the rest of the plants is unknown forms a basis for the planning a new investigation.

Ultimately, there must be unknown or unsure data for the planning of a new investigation.

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Answer: 360g potassium chlorate

40g KCIO3 - 400g H2O = 360g KCIO3

Answer: The entire container because gases will expand to fill it.

Explanation:

24.6 mL of distilled water must be added to 75.0 ml of 0.332 m FECL3(aq) to reduce its concentration to 0.250 m.

The abundance of a constituent divided by the total volume of a mixture is called concentration.

The reduction of the concentration from 0.332 M to 0.250 M corresponds to the dilution of : 0.332M/0.250 M = 1.328;

Add distilled water to the solution up to 1.328 times the first volume: 75 ml x 1.328 = 99.6 mL.

Now, take the 75 mL of 0.332 M, and add distilled water up to the volume of 99.6 mL, that is you add 24.6 mL of distilled water.

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If a sample contains 200 g of a radioactive isotope, 25 grams of the radioactive isotope will be left after 3 half-lives.

In this inquiry (t½) of isotope is 100 years, and that intends that following 100 years half of the example would have rotted and half would be left for what it's worth. 40 /2, or 20 g, decays after 100 years, or the first half life, and 20 g remains.

Formula used to calculate the number of half life elapsed :

N = N°/2ⁿ

where,

N = amount of radioactive sample remains after n-half lives = ?

= Initial amount of the radioactive sample= 10 g

n = number of half lives  = 3

N = 200g/2³

 = 25g

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If a sample contains 200 g of a radioactive isotope, 25 grams of the radioactive isotope will be left after 3 half-lives.

In this inquiry (t½) of isotope is 100 years, and that intends that following 100 years half of the example would have rotted and half would be left for what it's worth. 40 /2, or 20 g, decays after 100 years, or the first half life, and 20 g remains.

Formula used to calculate the number of half life elapsed :

N = N°/2ⁿ

where,

N = amount of radioactive sample remains after n-half lives = ?

= Initial amount of the radioactive sample= 10 g

n = number of half lives  = 3

N = 200g/2³

= 25g

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Nanomaterials, whether natural, engineered, organic, or inorganic, offer various applications across industries. However, their environmental health and safety impacts need to be carefully evaluated and managed to mitigate any potential risks.

Understanding their properties, fate, and behavior in different environments is crucial for responsible development, use, and disposal of nanomaterials.

Natural Nanomaterials:

Examples: Carbon nanotubes (CNTs) derived from natural sources like bamboo or cotton, silver nanoparticles in natural colloids, clay minerals (e.g., montmorillonite), iron oxide nanoparticles found in magnetite.

Uses: Natural nanomaterials have various applications in medicine, electronics, water treatment, energy storage, and environmental remediation.

Environmental health and safety impacts: The environmental impacts of natural nanomaterials can vary depending on their specific properties and applications. Concerns may arise regarding their potential toxicity, persistence in the environment, and possible accumulation in organisms. Proper disposal and regulation of their use are essential to minimize any adverse effects.

Engineered Nanomaterials:

Examples: Gold nanoparticles, quantum dots, titanium dioxide nanoparticles, carbon nanomaterials (e.g., graphene), silica nanoparticles.

Uses: Engineered nanomaterials have widespread applications in electronics, cosmetics, catalysis, energy storage, drug delivery systems, and sensors.

Environmental health and safety impacts: Engineered nanomaterials may pose potential risks to human health and the environment. Their small size and unique properties can lead to increased toxicity, bioaccumulation, and potential ecological disruptions. Safe handling, proper waste management, and risk assessment are necessary to mitigate any adverse effects.

Organic Nanomaterials:

Examples: Nanocellulose, dendrimers, liposomes, organic nanoparticles (e.g., polymeric nanoparticles), nanotubes made of organic polymers.

Uses: Organic nanomaterials find applications in drug delivery, tissue engineering, electronics, flexible displays, sensors, and optoelectronics.

Environmental health and safety impacts: The environmental impact of organic nanomaterials is still under investigation. Depending on their composition and properties, they may exhibit varying levels of biocompatibility and potential toxicity. Assessments of their environmental fate, exposure routes, and potential hazards are crucial for ensuring their safe use and minimizing any adverse effects.

Inorganic Nanomaterials:

Examples: Quantum dots (e.g., cadmium selenide), metal oxide nanoparticles (e.g., titanium dioxide), silver nanoparticles, magnetic nanoparticles (e.g., iron oxide), nanoscale zeolites.

Uses: Inorganic nanomaterials are utilized in electronics, catalysis, solar cells, water treatment, imaging, and antimicrobial applications.

Environmental health and safety impacts: Inorganic nanomaterials may have environmental impacts related to their potential toxicity, persistence, and release into ecosystems. Their interactions with living organisms and ecosystems require careful assessment to ensure their safe use and minimize any negative effects.

Understanding their properties, fate, and behavior in different environments is crucial for responsible development, use, and disposal of nanomaterials.

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Answer:

25lb 567.m

Explanation:

Answer:

the answer would be 11 grams of water/sugar water

Explanation: its that because 11 grams plus 1 gram = 11 grams

even though it may seem it may just go away that mass has to go somewhere it just doesn't vanish from the earth.

The number of protons and neutrons together makes the total atomic mass of the element. The atom of chromium has 28 neutrons. Thus, option b is correct.

An atomic mass is the property of an element that defines the number of protons and neutrons of the atom placed in a periodic table. The atomic mass is represented at the lower half of the atomic symbol.

The atomic mass is the sum of the neutrons and the protons that are held together in the nucleus of the atom as a concentrated mass. The atomic number is given as,

Atomic number = number of protons + number of neutrons

Given,

The atomic mass of chromium = 52

Number of protons = 24

Substituting values above:

52 = 24 + number of neutrons

number of neutrons = 52-24

= 28

Therefore, the number of neutrons of a chromium atom is 28.

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Answer: option b, 28

0.54g is the mass of carbon in  0.045 moles of carbon. Elementary particles shared the same quantity of matter.

A body's mass is an inherent attribute. Until the discoveries of the atom as well as particle physics, it was thought to be tied to the amount of matter inside a physical body. It was discovered that various atoms and elementary particles shared the same quantity of matter.

mole = given mass/ molar mass

substituting all the given values in the above equation, we get

0.045 moles = mass/ 12

mass =0.045×12= 0.54g

Therefore, 0.54g is the mass of carbon in  0.045 moles of carbon.

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Answer:

i think its a if not sorry i have it in a test right now

Explanation:

The 90% confidence interval for the difference in the incidence of hail between spraying and not spraying the clouds is (-0.141, 0.241).

The experiment aimed to determine if spraying chemicals on clouds for artificial rainfall could reduce hail occurrence. The data collected included the number of days when chemicals were sprayed on the clouds and the number of days when no intervention was done.

Out of the observed cloudy days with hail, 7 days had spraying while 43 days had no spraying. On the other hand, out of the observed cloudy days without hail, 43 days had spraying and 122 days had no spraying.

To calculate the 90% confidence interval for the difference in hail incidence, we use a formula that takes into account the sample sizes and proportions. After performing the calculations, the resulting confidence interval is (-0.141, 0.241). This means that there is not enough evidence to conclude a significant difference in hail occurrence between spraying and not spraying the clouds, as the confidence interval contains zero.

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5. To meet production in August, 71,000 pounds of raw materials are needed.

Since each unit of finished goods requires 5 pounds of raw materials, we can calculate the number of units needed in August as 71,000 pounds / 5 pounds per unit = 14,200 units.

The ending To determine the answers to the questions, we need to calculate the required information based on the provided data. inventory in July is 10% of the following month's raw materials production needs, which is 10% of 14,200 units = 1,420 units. Since each unit requires 5 pounds of raw materials, the total pounds of raw materials to be purchased in July is 1,420 units * 5 pounds per unit = 7,100 pounds. 6. To calculate the estimated cost of raw materials purchases for July, we need to determine the cost per pound of raw materials. Given that the raw materials cost $2.00 per pound, the estimated cost of raw materials purchases for July is 7,100 pounds * $2.00 per pound = $14,200. 7. The total estimated cash disbursements for raw materials purchases in July can be calculated by considering the payment terms. Twenty percent of raw materials purchases are paid for in the month of purchase, and 80% are paid in the following month. Given that the cost of raw material purchases in June is $93,040, the payment made in July is 20% * $93,040 = $18,608. The payment made in August would be 80% * $93,040 = $74,432. Therefore, the total estimated cash disbursements for raw materials purchases in July would be $18,608. 8. To calculate the estimated accounts payable balance at the end of July, we need to consider the payment terms. Since 80% of raw materials purchases are paid in the following month, 80% * $14,200 = $11,360 will be paid in August. Therefore, the estimated accounts payable balance at the end of July would be $14,200 - $11,360 = $2,840. 9. The estimated raw materials inventory balance at the end of July can be calculated by considering the ending raw materials inventory in July. Given that the ending raw materials inventory equals 10% of the following month's raw materials production needs, which is 10% of 14,200 units = 1,420 units. Since each unit requires 5 pounds of raw materials, the estimated raw materials inventory balance at the end of July would be 1,420 units * 5 pounds per unit = 7,100 pounds. 10. To calculate the total estimated direct labor cost for July, we need to consider the number of units produced and the direct labor wage rate. The budgeted unit sales for July are 12,000 units. Each unit of finished goods requires 2 direct labor-hours, so the total direct labor-hours for July would be 12,000 units * 2 direct labor-hours per unit = 24,000 direct labor-hours. Given that the direct labor wage rate is $13 per hour, the total estimated direct labor cost for July would be 24,000 direct labor-hours * $13 per hour = $312,000.

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5) 14,200 units of finished goods will be produced in August.

6) The estimated cost of raw materials purchases for July is  $2,800.

7) The total estimated cash disbursements for raw materials purchases in July is $18,608 + $74,432 = $93,040.

8) The estimated accounts payable balance at the end of July is  $59,546.40.

9)  Raw materials inventory balance is 700 pounds.

10) Total estimated direct labor cost for July is $312,000.

Let's solve  in detail:

5. To meet production in August, 71,000 pounds of raw materials are needed. Since each unit of finished goods requires 5 pounds of raw materials, the total number of units to be produced is 71,000 pounds / 5 pounds per unit = 14,200 units. Therefore, 14,200 units of finished goods will be produced in August.

6. If 7,000 pounds of raw materials are needed to meet production in August, and each unit of finished goods requires 5 pounds of raw materials, the total number of units to be produced is 7,000 pounds / 5 pounds per unit = 1,400 units.

The estimated cost of raw materials purchases for July is 1,400 units * $2.00 per pound = $2,800.

7. To calculate the total estimated cash disbursements for raw materials purchases in July, we need to consider the payment terms.

In July, 20% of raw materials purchases are paid for in the month of purchase, and 80% is paid in the following month.

The raw materials purchased in June amount to $93,040. Assuming 20% is paid in June, the cash disbursement for June is $93,040 * 20% = $18,608.

For the remaining 80% of the June purchases, which will be paid in July, we need to calculate the amount. This is given by 80% of the June purchases - 20% of the July purchases.

Amount to be paid in July = ($93,040 * 80%) - (Total purchases for July * 20%)

Solving for the Total purchases for July:

($93,040 * 80%) - (Total purchases for July * 20%) = Total purchases for July * 80%

Rearranging the equation:

($93,040 * 80%) = (Total purchases for July * 80% + Total purchases for July * 20%)

($93,040 * 80%) = Total purchases for July * 100%

Total purchases for July = ($93,040 * 80%) / 100%

Substituting the value of June purchases, we get:

Total purchases for July = ($93,040 * 80%) / 100% = $74,432

Therefore, the total estimated cash disbursements for raw materials purchases in July is $18,608 + $74,432 = $93,040.

8. If 7,000 pounds of raw materials are needed to meet production in August, and 80% of raw materials purchases are paid in the following month, the estimated accounts payable balance at the end of July would be 80% of the purchases for July.

Accounts payable balance = Total purchases for July * 80% = $74,432 * 80% = $59,546.40.

9. If 7,000 pounds of raw materials are needed to meet production in August, and the ending raw materials inventory equals 10% of the following month's raw materials production needs, the estimated raw materials inventory balance at the end of July would be 10% of the raw materials needed for August.

Raw materials inventory balance = 10% of the raw materials needed for August = 10% of 7,000 pounds = 700 pounds.

10. The total estimated direct labor cost for July can be calculated by multiplying the direct labor wage rate per hour by the total direct labor hours required for production in July.

Total direct labor hours for July = Number of units to be produced in July * Direct labor hours per unit = 12,000 units * 2 hours per unit = 24,000 hours.

Total estimated direct labor cost for July = Total direct labor hours for July * Direct labor wage rate = 24,000 hours * $13 per hour = $312,000.

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93.7 grams of copper(ii) oxide are produced when 1.00 kg of the copper ore is processed .

What is combustion ?

Combustion, a chemical reaction between substances, usually containing oxygen, that involves the production of heat and light, usually in the form of a flame. The speed or rate at which reactants combine is partly due to the nature of the chemical reaction itself, and partly due to the more energy produced than can escape to the surrounding medium, the temperature of the reactants is used to rise. further accelerate the reaction.

A well-known example of a combustion reaction is a lit match. When you light a match, the friction heats the head to a temperature at which the chemicals react, producing more heat than escaping into the air and burning in a flame.

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Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

\(P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ \)

From the question

actual volume = 4.13 mL

error = 4.13 - 4.09 = 0.04

We have

\(p(\%) = \frac{0.04}{4.13} \times 100 \\ = 0.968523002...\)

We have the final answer as

Hope this helps you

Answer:

Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)

The neutralization reaction is BHClO₃ + KOH → KClO₃ + H₂O. Acidic and basic qualities are often neutralized via neutralization processes, which combine an acid and a base to produce salt and water.

A strong acid and a strong base together will result in a neutral salt. When a strong acid and a weak base are combined, acid is created. Similar to this, when a weak acid is combined with a strong acid, a basic salt is created. There are several uses for neutralization.

In this reaction, a salt (KClO₃) and water (H₂O) are created when an acid (BHClO₃) combines with a base (KOH).

The neutralization reaction in the given options provided is:

BHClO₃ + KOH → KClO₃ + H₂O

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Answer:

0.25 liter

Explanation:

Trust

The percentage of aluminum in the gemstone is (2 x 26.98) / 537.49 x 100 = 10.04%.

The chemical formula of the emerald is Be3Al2Si6O18. , the percentage of aluminum in the gemstone is (2 x atomic weight of aluminum) / molecular weight of the compound x 100. The atomic weight of aluminum is 26.98 g/mol.
The molecular weight of the compound is (9 x atomic weight of Be) + (2 x atomic weight of Al) + (6 x atomic weight of Si) + (18 x atomic weight of O) = (9 x 9.01) + (2 x 26.98) + (6 x 28.09) + (18 x 16.00) = 537.49 g/mol.

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it's an ionic bond

btw i love your profile picture never understand is one of my favorites

Answer:

Explanation:

55 and 918

Answer:

The cone has a height of 10 and a diameter of 6. This means it has a radius of 3.

The volume of any cone is (1/3)πr2h. Therefore the volume of this cone would be (1/3)π(3)2(10) or 30π units3.

The hemisphere has a diameter of 6. This means it also has a radius of 3.

The volume of any sphere is (4/3)πr3. A hemisphere would be half that or (2/3)πr3. Therefore the volume of this hemisphere is (4/3)π(3)3 or 36π units3.The total volume would be 30π + 36π or 66π units3. The decimal approximation is 207.35 units3.

Explanation:

Cone: V = (1/3)πr2H

V = (1/3)π(3)2(10)

V = (1/3)π(90)

V = 30π

Hemisphere: V = (1/2)(4/3)πr3

V = (2/3)π(6)3

V = (2/3)π(216)

V = 144π

Total Volume = 30π + 144π = 174π

V ≈ 546.64 units3

Answer:

When moving from left to the right of a period, the number of electrons increases and the strength of shielding increases. As you move across period the number of shells remain same, the shielding effect will also remain constant.

Explanation:

The net ionic equation is written as H^+(aq) + OH^-(aq) -------> H2O(l)

The term net ionic equation refers to the equation that shows the ions that underwent a change in the reaction. We have to note that the reaction species here must be ionic species which are able to dissociate into ions in solutions.

Now the first step is to put down the molecular equation. The molecular equation shows the reaction of the compounds as follows;

HCI (aq) + NaOH (ag)-> NaCI (aq) +H2O(l)

Next, we put own the complete ionic reaction equation as follows;

H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) -------> Na^+(aq) + + Cl^-(aq) + H2O(l)

Next we have the net ionic equation;

H^+(aq) + OH^-(aq) -------> H2O(l)

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