A ______________________ is required to cause atoms to bond together or be separated from one another. *a physical change
b phase change
c radioactive decay
d chemical reaction

Answers

Answer 1

Answer:

Answer:

d chemical reaction..

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A metallic bond is formed between what two types of elements

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Answer:

metallic bonds are defined as those in which metals share valence electrons for example when sodium metallically bonds with itself each atom is sharing the electrons in the third orbital with up to eight other atoms the same thing happens with magnetism or other metals metallically bond to themselves.

How many significant digits are in 89015?

Answers

Answer:5

Explanation:

(e) Which sample contains more atoms: 2.0 g of Fe2O3 or 2.0 g of CaSO4? Support your answer with calculations.

Answers

CaSO4. It has 6 atoms while Fe2O3 has 5 atoms

Explanation:

Fe2O3 contains 2 atoms of Iron(Fe) and 3 atoms of Oxygen(O) = 2 + 3 = 5 atoms

CaSO4 contains 1 atom of Calcium(Ca), 1 atom of Sulphur(S) and 4 atoms of Oxygen(O) = 1 + 1 + 4 = 6 atoms

If 5.738 grams of AgNO3 is mixed with 4.115 grams of BaCl2 and allowed to react according to the balanced equation: BaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ba(NO3)2(aq) What is the limiting reagent? BaCl2AgNO3 How many grams of AgCl could be produced? grams AgCl What mass, in grams, of the excess reagent will remain? grams of excess reagent

Answers

Answer:

Limiting reagent: AgNO3

grams AgCl : 2.44 g AgCl

grams of excess reagent remain: 0.62 g BaCl2

Explanation:

1. Change grams to mol:

AgNO3:

5.738g x (1mol/169.87g) = 0.034 mol AgNO3

BaCl2:

4.115g x (1 mol/208.23g) = 0.020 mol BaCl2

2. Limiting reagent:

AgNO3:

0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2

BaCl2:

0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3

Limiting reagent: AgNO3

3. Grams of AgCl produced:

Using the limiting reagent:

0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl

4. Change mol to grams:

0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl

5. Grams of the excess reagent:

0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2

0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2

0.003 mol BaCl2 x ( 208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2

How many grams of sucrose must be added to 375 mL of watertoprepare a 2.75m/m percent solution of sucrose?

Answers

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 375 mL

$\text{Mass of water}=1.00g/mL* 375mL=375g$

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain $(375)/(97.25)* 2.75=10.6$ grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

Density of water ≈ 1 g/mL

Mass of water = Volume of water × Density of water

Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

Mass of sucrose = (Desired m/m percent / 100) × Total mass

Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))

Now, calculate:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - 0.0275)

Mass of sucrose ≈ (2.75 / 100) × (375 g) / 0.9725

Mass of sucrose ≈ (2.75 × 375 g) / 100 / 0.9725

Mass of sucrose ≈ (1031.25 g) / 0.9725

Mass of sucrose ≈ 1061.98 g

So, approximately 1062 grams of sucrose must be added to 375 mL of water to prepare a 2.75 m/m percent solution of sucrose.

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What is the density of iron if it crystallizes in a body-centered cubic unit cell with an edge length of 287 pm

Answers

Answer

Density = 7.87g/cm^3

Explanation:

Density is the ratio of mass of the given object to the volume of the object, in this question iron is the given object, then we make use of atomic number of iron

Given:

Length= 287pm = 287*10^-10cm

Atomic mass of Fe= 56.0u

Z=2(for body centered cubic unit cell)

Avogadro number (N 0)=6.022× 10^23

Density= ZM/a^3 × N

Where

Z= body centered cubic unit cell

Then substitute

N= Avogadro's number

a=Length

Density = (2× 56)/(287*10^-10cm)^3 × (6.022 × 10^23)

Density = 7.87g/cm^3

Final answer:

The density of iron in a body-centered cubic unit cell can be calculated using the mass and volume of the unit cell.

Explanation:

The density of iron can be calculated using the formula: density = mass/volume. To determine the mass of the unit cell, we need to know the molar mass of iron and the number of atoms in the unit cell. The molar mass of iron is 55.845 g/mol, and there are two iron atoms in the body-centered cubic unit cell of iron. The volume of the unit cell can be calculated using the formula: volume = (edge length)^3.

Putting these values into the formula, we get:

density = (2 * 55.845 g/mol) / ((287 pm)^3)

Converting the edge length to meters (1 pm = 1e-12 m) and calculating, we find that the density of iron is approximately 7.86 g/cm³.