CHEMISTRY COLLEGE

HELP Which type of light is stored energy?A. Kinetic
B. Potential
C. Thermal
D. Field

Answers

Answer 1
Answer:

Answer:

B

Explanation:
Answer 2
Answer:

Answer:

Potential

Explanation:

Related Questions

Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.
A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture
Write a balanced nuclear equation for the following: The nuclide uranium-238 undergoes alpha emission.
Exactly 1.5 g of a fuel burns under conditions of constant pressure and then again under conditions of constant volume. In measurement A the reaction produces 25.9 kJ of heat , and in measurement B the reaction produces 23.3 kJ of heat. Which measurement (A or B) corresponds to conditions of constant pressure? Which one corresponds to conditions of constant volume? Explain.
Write the reaction of ammonia and water. Label the acid, base, conjugate acid and conjugate base.Is Ammonia a weak or strong base?

Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make?

Answers

Answer:- 353.3 g

Solution:- The balanced equation is:

2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2

We start with given grams of calcium chloride and convert them to moles. Then using mol ratio, the moles of calcium phosphate are calculated and converted to grams as.

Molar mass of calcium chloride is 110.98 gram per mol and molar mass of calcium phosphate is 310 gram per mol.

The set is made using dimensional analysis as:

379.4gCaCl_2((1molCaCl_2)/(110.98gCaCl_2))((1molCa_3(PO_4)_2)/(3molCaCl_2))((310gCa_3(PO_4)_2)/(1molCa_3(PO_4)_2))

= 353.3gCa_3(PO_4)_2

So, 353.3 grams of calcium phosphate can be formed.

Answer:

353.3g

Explanation:

When radioactive uranium decays to produce thorium, it also emits a particle. As seen in the balanced nuclear equation, this particle can BEST be described asA) a helium atom.
B) an alpha particle or a helium atom.
C) a beta particle or a hydrogen nucleus.
D) an alpha particle or a helium nucleus.

Answers

The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.

What is radioactive decay?

Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.

Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,

In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.

To find more on alpha decay, refer here:

brainly.com/question/27870937

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A volume of 40.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2SO4). What was the molarity of the KOHKOH solution if 16.2 mLmL of 1.50 MM H2SO4H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Answers

Answer: The concentration of KOH solution is 1.215 M

Explanation:

For the given chemical equation:

2KOH(aq.)+H_2SO_4(aq.)\rightarrow K_2SO_4(aq.)+2H_2O(l)

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\nM_1=1.50M\nV_1=16.2mL\nn_2=1\nM_2=?M\nV_2=40.0mL

Putting values in above equation, we get:

2* 1.50* 16.2=1* M_2* 40.0\n\nM_2=(2* 1.50* 16.2)/(1* 40.00)=1.215M

Hence, the concentration of KOH solution is 1.215 M

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?​

Answers

Answer:

Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^(+2)(aq)+2(PO_4)^(-3)(aq)

Ksp=[Sr^(+2)]^3[(PO_4)^(-3)]^2

Explanation:

Hello,

In this case, for strontium phosphate, we find an ionic equation for its dissociation as shown below:

Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^(+2)(aq)+2(PO_4)^(-3)(aq)

Next, the solubility product is found by applying the law of mass action, considering that the solid salt is not considered but just the aqueous species due to heterogeneous equilibrium:

Ksp=[Sr^(+2)]^3[(PO_4)^(-3)]^2

Best regards.

Which compound is classified as an electrolyte? a) C6H12O6
b)Ca(OH)2
c)C12H22O11
d)CH3OH

Answers

Answer:
            Option-B, Ca(OH)₂ is the correct answer.

Explanation:
                   By definition Electrolytes are those compounds which when poured in polar solvents dissociate into a positively charged specie called Cation and a negatively charged specie called an Anion. These ions have the potential to conduct electricity. 
                   So Ca(OH)₂ is dissolved in water it produces Ca⁺² cation and OH⁻ anions. So it acts as an Electrolyte.
                    While, C₆H₁₂O₆ (Glucose), C₁₂H₂₂O₁₁ (Table Sugar) and CH₃OH (Methanol) are covalent compounds and does not produce any positive or negative ions when dissolved in water.

An alloy is a mixture of metals to obtain a more durable and strongersubstance.
True
False

Answers

Answer: False

Explanation: Alloys are harder and stronger because the different-sized atoms of the mixed metals make the atomic layers less regular, so they cannot slide as easily.
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Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mLmL of water to produce a solution that freezes at −−14.5 ∘C? The freezing point for pure water is 0.0 ∘C∘C and Kf is equal to 1.86 ∘C/m.If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.