Question 1
1 pts
How many mols of bromine are present in 35.7g of
Tin(IV) bromate?

Answer:

pH - 3.41 = acidic

pH = 10.25 = basic

pH = 7.00 = neutral

[H+] -3.5 x 10-5 = acidic

[H+] - 6.7 x 10-9 = basic

[OH-]-5.8 x 10-4 = basic

[H0] -1.0 x 10-7 = neutral

[OH-] - 4.5 x 10-13 = acidic

Explanation:

Let us note that from the pH scale, a pH of;

0 - 6.9 is acidic

7 is neutral

8 - 14 is basic

But pH= - log [H^+]

pOH = -log [OH^-]

Then;

pH + pOH = 14

Hence;

pH = 14 - pOH

For [H+] -3.5 x 10-5

pH = 4.46 hence it is acidic

For [H+] - 6.7 x 10-9

pH = 8.17 hence it is basic

[OH-]-5.8 x 10-4

pH= 10.76 hence it is basic

[H0] -1.0 x 10-7

pH = 7 hence it is neutral

[OH-] - 4.5 x 10-13

pH = 1.65 hence it is acidic

Answer:

Calculate the moles of N2 molecules in 3.94 grams of nitrogen.

Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules.

Explanation:

Calculate the moles of N2 molecules in 4.73 liters of nitrogen gas. FALSE. You can't make this conversion using only the conversion factor with units of g/mol. To convert liters to moles are necessaries pressure, temperature and volume of the gas to use PV = nRT

Calculate the grams of N2 in 10.58 liters of nitrogen gas. FALSE. As explained, you need, P,V and T to find the moles of the gas. With the moles you can find the mass using the conversion factor of 28.02g/mol

Calculate the moles of N2 molecules in 3.94 grams of nitrogen. TRUE. You can find the moles of N2 as follows:

3.94g N2 * (1mol/28.02g) = 0.14 moles of N2 molecules

Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules. TRUE. The mass in 5.03x10²⁰ moles of nitrogen molecules is:

5.03x10²⁰ moles * (28.02g/mol) = 1.4x10²²g of nitrogen.

Answer:

10mL

Explanation:

Using the formula as follows:

CaVa = CbVb

Where;

Ca = concentration of acid, HBr (M)

Cb = concentration of base, Sr(OH)2 (M)

Va = volume of acid, HBr (Litres)

Vb = volume of base, Sr(OH)2 (Litres)

According to the information given in this question;

Ca = 0.24M

Cb = 0.24M

Va = ?

Vb = 10.0ml

Using CaVa = CbVb

0.24 × Va = 0.24 × 10

0.24Va = 2.4

Va = 2.4 ÷ 0.24

Va = 10mL

10mL of HBr is needed.

Answer:

11

1. 6.02×10 23

this is the answer Hope it helps you

Answer:

Bohr Model is the correct answer

Answer:

Electron -Cloud Model

Explanation:

Just took the quiz got 100%

Answer: A volume of 37.5 mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water.

Explanation:

Given: [tex]M_{1}[/tex] = 0.300 M,   [tex]V_{1}[/tex] = ?

[tex]M_{2}[/tex] = 0.0450 M,    [tex]V_{1}[/tex] = 250.0 mL

Formula used is as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

Substitute the values into above formula as follows.

[tex]M_{1}V_{1} = M_{2}V_{2} \\0.300 M \times V_{1} = 0.0450 M \times 250.0 mL\\V_{1} = 37.5 mL[/tex]

Thus, we can conclude that a volume of 37.5 mL of 0.300 M NaF would be required to make a 0.0450 M solution of NaF when diluted to 250.0 mL with water.

Explanation:

Population density is defined as the number of people present per square kilometre. Population density of India according to 2011 census is 382 persons per square kilometres.

Answer:

Option A (0.043 g) is the correct answer.

Explanation:

Given:

= 43 mg

As we know,

[tex]1 \ mg = \frac{1}{1000} \ g[/tex]

then,

⇒ [tex]43 \ mg = \frac{43}{1000} \ g[/tex]

              [tex]= 0.043 \ g[/tex]

Thus, the above is the correct alternative.

Answer:

Ni

Explanation:

An active metal is a highly reactive metal. Active metals are found high up in the activity series.

Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.

Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.

Answer:

N 2 H4

Explanation:

the answer to the question is B.UV-B

Answer:

1) 16.0 moles Al

24.0 moles S

96.0 moles O

2)In 6.10 moles magnesium perchlorate, (Mg(CIO4)2 we have:

6.10 moles Mg

12.2 moles Cl

48.8 moles O

3)4.6 moles of propane (total) contains 13.8 moles of carbon and 36.8 moles of hydrogen atoms

4)The gold coin contains 7.8 *10^22 atoms

Explanation:

Step 1: Data given

Number of moles of aluminum sulfate, Al2(SO4)3 = 8.00 moles

Step 2: Calculate the number of moles

In 1 mol of aluminum sulfate, Al2(SO4)3 we have:

2 moles of Al

3 moles of S

12 moles of O

This means that in 8.00 moles of aluminum sulfate, Al2(SO4)3 we have:

2*8.00 = 16.0 moles Al

3*8.00 = 24.0 moles S

12*8 = 96.0 moles O

2. Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 6.10 moles of magnesium perchlorate, (Mg(CIO4)2

1 mol of magnesium perchlorate, (Mg(CIO4)2 has:

1 Mol of Mg

2 moles of Cl

8 moles of O

In 6.10 moles magnesium perchlorate, (Mg(CIO4)2 we have:

1 * 6.10 moles = 6.10 moles Mg

2*6.10 = 12.2 moles Cl

8*6.10 = 48.8 moles O

3. A sample of propane, C3H8, contains 13.8 moles of carbon atoms. How many total moles of atoms does the sample contain?

In 1 mol of propane, C3H8 we have:

3 moles of C and 8 moles of H

This means if we have 13.8 moles of carbon, we have 13.8/3 = 4.6 moles of propane, C3H8 and 4.6 *8 = 36.8 moles of H

So 4.6 moles of propane contains 13.8 moles of carbon and 36.8 moles of hydrogen atoms

4. A rare gold coin (24 karat, or 100% gold) has a mass of 25.54 g. How many atoms of gold are present in this coin?

Calculate moles of gold:

Moles = mass of gold / molar mass gold

Moles = 25.54 grams / 196.97 g/mol

Moles = 0.1297 moles

Calculate atoms:

Number of atoms = moles * number of Avogadro

0.1297 * 6.022 *10^23 = 7.8 *10^22 atoms

The gold coin contains 7.8 *10^22 atoms

Answer:

9.36 g

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)

Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles

1 mole of octane yields 9 moles of water

0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water

Number of moles of oxygen = 23g/32g/mol = 0.72 moles

12.5 moles of oxygen yields 9 moles of water

0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water

Hence oxygen is the limiting reactant;

Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g

Answer:

Sterilization refers to any process that removes, kills, or deactivates all forms of life and other biological agents like prions present in a specific surface, object or fluid, for example food or biological culture media.

Each triglyceride is different from the others on the basis of presence of fatty acids in it.

Triglycerides is a kind of fat and derivative of ester which is formed by the combination of glycerol and three fatty acids.

So in the triglyceride molecule three sub divided parts are present due to the presence of three fatty acids groups and these fatty acids will make difference in each triglyceride molecules.

Hence of fatty acids in triglyceride molecule makes it different from other.

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Answer:

It has different fatty acids.

Explanation:

This is a signature of triglycerides

Answer:

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

Explanation:

Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂

Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)

The complete ionic equation includes all the ions and insoluble species.

Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)

The net ionic equation includes only the ions that participate in the reaction and insoluble species.

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

Answer:

c

Explanation:

its c because it has multiple mixture blueberries flower water and others thats why i says c

Answer:

Overshooting the endpoint leads to a percent acetic acid in vinegar higher than the correct value.

Explanation:

When too much NaOH solution is added, the resulting number of equivalent hydronium (OH⁻) will be higher than what it is in reality. This would directly lead to the number of acetic acid moles in the vinegar being found higher than normal.

Answer:

0.111 g

Explanation:

1 g = 1000 mg

Doctor ordered the following concentration of Claforan:

C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL

If we add 2 g iof Claforan, we obtain:

2 g Claforn ---- 180 mg/mL Claforan

To reach a concentration equal to C (10 mg/mL), we need:

10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn

Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.  

Answer:

It would take 200,000 years for a spaceship traveling at the speed of light to go across the entire galaxy.

Answer:

1.40x10^24 atoms of H

Explanation:

Answer:

A positively charged object will exert a repulsive force upon a second positively charged. This repulsive force will push the two objects apart while a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other

The interaction between objects with positive and negative electric charges can be analogously modeled using magnets. The Types of Forces Involved are; Attractive Magnetic Force, Repulsive Magnetic Force and the Energy Transformations are; Potential Energy Transformation, and Kinetic Energy Transformation.

In this analogy, magnets can represent the charges, and magnetic forces can represent the electric forces.

Interaction Between Magnets

Imagine we have two magnets: one with a north pole (N) and the other with a south pole (S). When you bring the north pole of one magnet close to the south pole of the other magnet, they are attracted to each other. Conversely, if you bring the north pole of one magnet near the north pole of the other magnet, they repel each other.

Types of Forces Involved:

Attractive Magnetic Force (Analogous to Electric Attraction):

When the north pole of one magnet is brought close to the south pole of another magnet, they experience an attractive magnetic force. Similarly, when objects with opposite electric charges were brought close together, then they will experience an attractive electric force.

Repulsive Magnetic Force (Analogous to Electric Repulsion):

When two magnets with the same pole (both north or both south) are brought close to each other, they experience a repulsive magnetic force. This is analogous to the repulsion between objects with like electric charges (both positive or both negative).

Energy Transformations;

When you bring the magnets closer together or move them apart, energy transformations occur:

Potential Energy Transformation;

As the magnets are moved closer together, the potential energy of the magnetic interaction decreases. This is because the magnets' magnetic fields interact more strongly, and they tend to move toward each other due to the attractive or repulsive forces.

Kinetic Energy Transformation;

If you let the magnets go after bringing them close together, they will move towards each other (in the case of attraction) or move apart (in the case of repulsion). This movement involves a transformation of potential energy into kinetic energy. The kinetic energy increases as the magnets move, and it's at its maximum when the magnets are farthest apart (in the case of repulsion) or when they collide (in the case of attraction).

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Answer:

Oxygen, O₂ is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2H₂ + O₂ —> 2H₂O

Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂O from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Therefore, 10 g of H₂ will react with

= (10 × 32)/4 = 80 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.

Oxygen has been the limiting reactant in the reaction.

A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.

The balanced equation for the formation of water has been:

[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]

For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.

The moles can be calculated as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:

Moles = [tex]\rm \dfrac{10}{2}[/tex]

Moles of hydrogen = 5 mol.

Moles of Oxygen in 5 grams Oxygen:

Moles = [tex]\rm \dfrac{5}{32}[/tex]

Moles of oxygen = 0.156 mol.

For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.

For reacting with 5 mol of Hydrogen, moles of oxygen required are:

Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]

Moles of oxygen required = 2.5 moles.

The available oxygen = 0.156 moles.

Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.

Thus, oxygen has been the limiting reactant in the reaction.

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ummm is that chemistry?

Answer:

is this chem

Explanation:

Answer:

Volume of hot water required = 3.85L

Explanation:

Suppose volume of hot Then volume of water required cold water = = x L (10.0-x) L

Heat given by hot water (Q₁)

= mass of hot water x heat capacity of water X AT

= x L * 4.184 * J / g. к x(61.0-37.0) °℃.

And Heat absorbed by cold water (Q₂) = (10.0-x) L x 4.184 J/g*k x(37+0 -220) C

Since energy is consumed, Q₁ = Q2.

i.e. X*l *4.184*J/g*k*24C = (10.0-x)L x 184 5

24 x 15 (10.0-x) = 150. - 15x

x = 150. (24+15) = 3.846

So, volume of hot water required. = 3.85 L

When the temperature of the water increases the water becomes hot.

According to the question the volume of hot water required = 3.85L.

Suppose volume of hot Then the volume of water required cold water  is [tex]x L (10.0-x) L[/tex]

All the data are given in the question, which is as follows:-

The formula we are going to use is as follows:-

= mass of hot water x heat capacity of water X AT

= [tex]x L * 4.184 *(61.0-37.0) ^oC[/tex]

The heat absorbed by cold water (Q₂) = [tex](10.0-x) L *4.184 *(37+0 -220) ^oC[/tex]

Since energy is consumed, Q₁ = Q2.

[tex]X*l *4.18424C = (10.0-x)L * 184 524 * 15 (10.0-x) = 150. - 15xx = 150. (24+15) = 3.846[/tex]

Hence, the volume of hot water required is = 3.85 L

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The reduction of the species defines the gain of electrons. The iron is most likely to be reduced when reacts with zinc. Thus option A is correct.

Oxidizing agents are the species that gain electrons and get reduced, their oxidation number gets reduced when the metal reacts.

In the reactivity, series zinc is placed before iron and hence is a reducing agent that gets oxidized. Down the series, the reducing ability decreases while the oxidizing increases.

Therefore, option A. iron will be reduced when reacts with zinc.

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