How do the causes of surface and deep water currents differ?
A. Surface currents are caused by wind deep water currents are caused by difference in water density
B. Surface currents are caused by the Coriolis effect deep water currents are caused by differences in water density
C. Surface currents are caused by differences in water salinity deep water currents are caused by differences in water temperature
D. Surface currents are caused by differences in water density deep water currents are caused by wind

Answer:

Due to the circulation of the wind and water the trash will continue to go around in a circle thus causing it to be trapped there.

Solution :

The reasons for the melting point determines  :

In order to prepare the sample in the capillary tube for determining its melting point, we tap the tube gently into the sample with the open end of tube down. We then continue tapping the tube till the sample is couple of millimeters high.

Then the other close end of tube down, we tap the sample down again slowly or then dropping the tube into a longer tube so as to move down the sample faster.

Answer:

176 grams of sugar

explanation is down

Explanation:

250 / 22 = 2000 / x250 ml to 22 grams = 2000 ml to x grams

cross multiply

(250)(x) = (2000)(22)

250x = 44,000

x = 44,000/250

x = 176 grams of sugar 

Answer: The value of the equilibrium constant Kc for this reaction is 0.088

Explanation:

[tex]Molarity=\frac{x}{M\times V_s}[/tex]

where,

x = given mass

M = molar mass

[tex]V_s[/tex] = volume of solution in L

Equilibrium concentration of [tex]CaCO_3[/tex] = [tex]\frac{25.3}{100\times 9.0}=0.028M[/tex]

Equilibrium concentration of [tex]CaO[/tex] = [tex]\frac{14.9}{56\times 9.0}=0.029M[/tex]

Equilibrium concentration of [tex]CO_2[/tex] = [tex]\frac{33.7}{44\times 9.0}=0.085M[/tex]

The given balanced equilibrium reaction is,

[tex]CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)[/tex]

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CaO]\times [CO_2]}{[CaCO_3]}[/tex]  

Now put all the given values in this expression, we get :

[tex]K_c=\frac{0.029\times 0.085}{0.028}=0.088[/tex]

5.55 mol C₂H₅OH

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

Step 2: Identify Conversions

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

Step 3: Stoichiometry

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

Answer:

Ball 16 oz Pint Mason Jar outer dimensions are 5.2 height by 3.2 diameter – which could hold approximately 390 jelly beans. Ball 32 oz Quart Mason Jar outer dimensions are 6.9 height by 3.9 diameter – which could hold approximately 804 jelly beans.Oct 5, 2017

Explanation:

please mark me as brainliest

Answer:

100 jellybeans

Explanation:

Mass of Jar with jellybeans=230 grams

Mass of Jar=30 grams

Mass of 1 jellybean=2 grams

Mass of Jellybeans in jar

=Mass of Jar with jellybeans-Mass of Jar

=230 grams - 30 grams

=200 grams

2 grams=1 jellybean

1 gram=1/2 jellybean

200 grams=200*(1/2) jellybeans=100 jellybeans

Answer:

11. Polonium.

12. Argon.

Explanation:

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

11. Polonium: this metalloid has 6 valence electrons and 6 electron shells. It has 84 electrons with 6 shell structure 2, 8, 18, 32, 18, 6.

Therefore, polonium has the following electronic configuration;

Polonium = 1s² 2s² 2p² 3s² 3p6 3d10 4s² 4p6 4d10 5s² 5p6 4f¹⁴ 5d10 6s² 6p⁴

12. Argon: this element has the same number of electrons as a Sulfur ion. It has 18 electrons.

Answer:

T = 4.062V

Explanation:

from PV = nRT => T = PV/RT

P = 1 atm

V = Final Volume

n = 3 moles

R = 0.08206 L·atm/mol·K

T = ?

T = 1 atm · V(Liters)/(3 moles)(0.08206L·atm/mol·K) = 4.062·V(final) Kelvin

The temperature of the balloon is 406 K

We'll begin by listing out what was given from the question. This is shown below:

Volume (V) = 100 L

Mole of oxygen (n) = 3 moles

Pressure (P) = 1 atm

We can obtain the temperature of the balloon by using the ideal gas equation as shown below:

P is the pressure.

V is the volume.

n is the number of mole

R is the gas constant (0.0821 atm.L/Kmol)

T is the temperature.

Applying the ideal gas equation, we have:

PV = nRT

1 × 100 = 3 × 0.0821 × T

100 = 0.2463 × T

Divide both side by 0.2463

T = 100 / 0.2463

Therefore, the temperature of the balloon is 406 K

Learn more: https://brainly.com/question/17046345

Answer:

im only in 6th grade TwT

Explanation:

LOL

Answer:

Yes

Explanation:

The point of greenhouse gases is that it absorbs and emits radiant energy within the thermal infrared range and causes it to be trapped.

Answer:

Zn

Explanation:

Answer:

1.86 atm

Explanation:

The following data were obtained from the question:

Initial volume (V₁) = 18.6 L

Initial pressure (P₁) = 0.10 atm

Final volume (V₂) = 1000 mL

Final pressure (P₂) =?

NOTE: The temperature is constant.

Next, we shall convert 1000 mL to L.

1000 mL = 1 L

Finally, we shall determine the final pressure of the gas as follow:

Initial volume (V₁) = 18.6 L

Initial pressure (P₁) = 0.10 atm

Final volume (V₂) = 1 L

Final pressure (P₂) =?

P₁V₁ = P₂V₂

18.6 × 0.10 = P₂ × 1

1.86 = P₂

P₂ = 1.86 atm

Thus, the final pressure of the gas is 1.86 atm

Answer:

In 1851 gold-seekers from around the world began pouring into the colonies, changing the course of Australian history. The gold rushes greatly expanded Australia's population, boosted its economy, and led to the emergence of a new national identity.

Explanation:

Answer:

People from all over the world look for gold and in 1851 people started forming colonies to look for it which had a huge impact on the history of Australia.   The gold rushes increased Australia's population which improved its economy.

Answer:

T₁ = P₁V₁T₂ / P₂V₂

Explanation:

The combine gas equation is given by:

P₁V₁ / T₁ = P₂V₂ / T₂

Where:

P₁ => is the initial pressure

V₁ => is the initial volume

T₁ => is the initial temperature

P₂ => is the final pressure

V₂ => is the final volume

T₂ => is the final temperature

Finally, we shall make T₁ the subject of the above expression. This can be obtained as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

Cross multiply

T₁P₂V₂ = P₁V₁T₂

Divide both side by P₂V₂

T₁ = P₁V₁T₂ / P₂V₂

Thus, the formula for T₁ is:

T₁ = P₁V₁T₂ / P₂V₂

Answer:

pH= 1.17

Explanation:

The neutralization reaction between HBr (acid) and KOH (base) is given by the following equation:

HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)

According to this equation, 1 mol of HBr reacts with 1 mol of KOH. Then, the moles can be expressed as the product between the molarity of the acid/base solution (M) and the volume in liters (V). So, we calculate the moles of acid and base:

Acid:

M(HBr) = 0.15 M = 0.15 mol/L

V(HBr) = 50.0 mL x 1 L/1000 mL = 0.05 L

moles of HBr = M(HBr) x V(HBr) = 0.15 mol/L x 0.05 L = 7.5 x 10⁻³ moles HBr

Base:

M(KOH) = 0.25 M = 0.25 mol/L

V(HBr) = 13.0 mL x 1 L/1000 mL = 0.013 L

moles of HBr = M(HBr) x V(HBr) = 0.25 mol/L x 0.013 L = 3.25 x 10⁻³ moles KOH

Now, we have: 7.5 x 10⁻³ moles HBr > 3.25 x 10⁻³ moles KOH

HBr is a strong acid and KOH is a strong base, so they are completely dissociated in water: the acid produces H⁺ ions and the base produces OH⁻ ions. So, the difference between the moles of HBr and the moles of KOH is equal to the moles of remaining H⁺ ions after neutralization:

moles of H⁺ = 7.5 x 10⁻³ moles HBr - 3.25 x 10⁻³ moles KOH = 4.25 x 10⁻³ moles H⁺

From the definition of pH:

pH = -log [H⁺]

The concentration of H⁺ ions is calculated from the moles of H⁺ divided into the total volume:

total volume = V(HBr) + V(KOH) = 0.05 L + 0.013 L = 0.063 L

[H⁺] = (moles of H⁺)/(total volume) = 4.25 x 10⁻³ moles/0.063 L = 0.067 M

Finally, we calculate the pH after neutralization:

pH = -log [H⁺] = -log (0.067) = 1.17

Solution :

Partition coefficient Kd

[tex]$=\frac{\frac{\text{mass of caffeine in } CH_2Cl_2}{\text{volume of } CH_2Cl_2}}{\frac{\text{mass of caffeine in water}}{\text{volume of water}}}$[/tex]

= 9.0

A). 1 x 200 mL extraction :

Let m be the mass of caffeine in the water

Mass of caffeine in [tex]$CH_2Cl_2$[/tex] = 100 - m

[tex]$\frac{\frac{100-m}{200}}{\frac{m}{200}} = 9.0$[/tex]

[tex]$=\frac{100-m}{m} = 9.0$[/tex]

[tex]$= 10 m = 100$[/tex]

m = 10

Mass remaining in coffee = m =  10 mg

B). 2 x 100 mL extraction :

First extraction :

Let [tex]$m_1$[/tex] be the mass of caffeine in the water

Mass of caffeine in [tex]$CH_2Cl_2$[/tex] = [tex]$100-m_1$[/tex]

[tex]$\frac{\frac{100-m_1}{100}}{\frac{m_1}{200}} = 9.0$[/tex]

[tex]$=\frac{100-m_1}{m_1} = 4.5$[/tex]

[tex]$ 5.5 \ m_1 = 100$[/tex]

[tex]$m_1$[/tex]  = 18.18

Mass remaining in coffee = [tex]$m_1$[/tex] =  18.18 mg

Second Extraction :

Let [tex]$m_2$[/tex] be the mass of caffeine in the water

Mass of caffeine in [tex]$CH_2Cl_2$[/tex] = [tex]$18.18-m_2$[/tex]

[tex]$\frac{\frac{18.18-m_2}{100}}{\frac{m_2}{200}} = 9.0$[/tex]

[tex]$=\frac{18.18-m_2}{m_2} = 4.5$[/tex]

[tex]$ 5.5 \ m_2 = 18.18$[/tex]

[tex]$m_2$[/tex]  = 3.3

Mass remaining in coffee = [tex]$m_2$[/tex] =  3.3 mg

Answer:

C

Explanation:

Wind is geological therefore it is geological weathering

Answer:

C. Carbon monoxide

Explanation:

Chemical substance which has a triple covalent bond is carbon monoxide.

Covalent bond is defined as a type of bond which is formed by the mutual sharing of electrons to form electron pairs between the two atoms.These electron pairs are called as bonding pairs or shared pair of electrons.

Due to the sharing of valence electrons , the atoms are able to achieve a stable electronic configuration . Covalent bonding involves many types of interactions like σ bonding,π bonding ,metal-to-metal bonding ,etc.

Sigma bonds are the strongest covalent bonds while the pi bonds are weaker covalent bonds .Covalent bonds are affected by electronegativities of the atoms present in the molecules.Compounds having covalent bonds have lower melting points as compared to those with ionic bonds.

Learn more about covalent bonds,here:

https://brainly.com/question/19382448

#SPJ2

Answer:

A.

Explanation:

A decomposition reaction is when one reactant (the substances to the left of the arrow in the reaction) breaks apart into two or more products (the substances to the right of the arrow).

Because KClO3 is breaking down into KCl and O2, it's a decomposition reaction.

Answer:

Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO2). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.

Explanation:

Please give me brainlist

Answer:

again, I'm going to say D lol.

Answer:

K = 8.6x10⁻⁶

Explanation:

a chemist finds that a 5.2L reaction vessel...

To solve this question we need first to find the equation of the equilibrium constant using the chemical eqiation:

TiCl₄(l) ⇄ Ti(s) + 2Cl₂(g)

The equilibrium constant expression is:

K = [Cl₂]²

Because equilibrium constant is defined as the ratio berween concentrationa of products over reactant powered to its reaction coefficient. But pure liquids as TiCl₄(l) and pure solids as Ti(s) are not taken into account

Now, we need to find the molar concentration of Cl₂, [Cl₂]:

Moles Cl₂ -Molar mass: 70.9g/mol-:

1.08g * (1mol / 70.9g) = 0.0152 moles / 5.2L =

2.93x10⁻³M = [Cl₂]

K = (2.93x10⁻³)²

The question is incomplete, the complete question is;

Draw Lewis structures for the following compounds/ions.

Decide if the compound/ion listed can satisfy the octet rule (duet rule for H), if so choose Satisfies Octet Rule from the drop down menu.

If the compound/ion cannot satisfy the octet rule for one of the atoms in the formula, choose one of the following three reasons why;

Too many valence electrons (more than eight), Too few valence electrons (fewer than eight), or Odd # of valence electrons.

Satisfies Octet Rule Too many valence electrons Too few valence electrons Odd # of valence electrons  KrCl2

Satisfies Octet Rule Too many valence electrons Too few valence electrons Odd # of valence electrons  PCl3

Satisfies Octet Rule Too many valence electrons Too few valence electrons Odd # of valence electrons  BeH2

Satisfies Octet Rule Too many valence electrons Too few valence electrons Odd # of valence electrons  NO3-

Satisfies Octet Rule Too many valence electrons Too few valence electrons Odd # of valence electrons  SeO22-

Satisfies Octet Rule Too many valence electrons Too few valence electrons Odd # of valence electrons  SO2

Answer:

See explanation

Explanation:

Let us consider the number of valence electrons in each of the species. This will let us know whether the specie satisfies the octet rule or not.

KrCl2 has a total of 22 valence electrons. This implies that there are too many valence electrons hence the compound does not satisfy the octet rule.

For PCl3, there are 26 valence electrons. This compound satisfies the octet rule.

For BeH2, there are four valence electrons. This number of valence electrons is too few hence BeH2 does not satisfy the octet rule.

NO3^- has 24 valence electrons and is represented using resonance structures. Its resonance structures show that the specie satisfies the octet rule.

SeO2^2- has 20 valence electrons. This specie satisfies the octet rule.

SO2 has 18 valence electrons and is represented by resonance structures which satisfy the octet rule.

Answer:

When an alkyl benzene is heated with strong oxidizing asgents like acidic or alkline KMnO4

or acidified K2Cr2O7

, etc. gives aromatic carboxyllic acid. The alkyl side chain gets oxidised to −COOH

group irrespective of the size of the chain.

Explanation: