If you know, answer this. If you don’t know this, don’t second guess

Answer: i think its the thired one

Explanation:

Answer:

again, I'm going to say D lol.

Answer:

The answer is A

Explanation:

I got it right on the test

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:

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:

A. 0.027 M

B. 0.0054 M

C. 31 cm³

Explanation:

A. Determination of the final concentration (Molarity)

Initial Volume (V₁) = 27 cm³

Initial concentration (C₁) = 0.150 M

Final volume (V₂) = 150 cm³

Final congratulation (C₂) =?

C₁V₁ = C₂V₂

0.150 × 27 = C₂ × 150

4.05 = C₂ × 150

Divide both side by 150

C₂ = 4.05 / 150

C₂ = 0.027 M

Thus, the final concentration of the solution is 0.027 M

B. Determination of the final concentration (Molarity)

Initial Volume (V₁) = 35.71 cm³

Initial concentration (C₁) = 0.0756 M

Final volume (V₂) = 500 cm³

Final congratulation (C₂) =?

C₁V₁ = C₂V₂

0.0756 × 35.71 = C₂ × 500

Divide both side by 500

C₂ = (0.0756 × 35.71) / 500

C₂ = 0.0054 M

Thus, the final concentration of the solution is 0.0054 M

C. Determination of the final volume.

Initial Volume (V₁) = 10 cm³

Initial concentration (C₁) = 0.155 M

Final congratulation (C₂) = 0.05 M

Final volume (V₂) =?

C₁V₁ = C₂V₂

0.155 × 10 = 0.05 × V₂

1.55 = 0.05 × V₂

Divide both side by 0.05

V₂ = 1.55 / 0.05

V₂ = 31 cm³

Thus, the final volume of the solution is 31 cm³

Answer:

its the last one

Explanation:

Answer:

Zn

Explanation:

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:

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: 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]

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 

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:

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:

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

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:

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:

1.  H₂SO₄  +  Na₂CO₃  ⇒  Na₂SO₄  +  H₂O  +  CO₂

sulfuric acid + sodium carbonate ⇒ sodium sulfate + water + carbon dioxide

2.  2 HCl  +  Mg(HCO₃)₂  ⇒  MgCl₂  +  2 H₂O  +  2 CO₂

hydrochloric acid + magnesium hydrogencarbonate ⇒ magnesium chloride + water + carbon dioxide

Explanation:

1.  Sulfuric acid is H₂SO₄.  Sodium carbonate is Na₂CO₃.

H₂SO₄  +  Na₂CO₃  ⇒  Na₂SO₄  +  H₂O  +  CO₂

sulfuric acid + sodium carbonate ⇒ sodium sulfate + water + carbon dioxide

2.  Hydrochloric acid is HCl.  Magnesium hydrogencarbonate is Mg(HCO₃)₂.

2 HCl  +  Mg(HCO₃)₂  ⇒  MgCl₂  +  2 H₂O  +  2 CO₂

hydrochloric acid + magnesium hydrogencarbonate ⇒ magnesium chloride + water + carbon dioxide

Answer:

should be D. Hope this helpsss!

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:

It can be verified since the information came from a government article. Or you could trust the magzine/general article if its from a well known article. Explanation:

you get the point, you can add on to this. I can't make it the best since im not in your class learning. Good luck

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

Answer:

2.44 g S

Explanation:

One mole of Na2SO4 contains 1 mol of S.

Step 1: Find out the moles of Na2SO4:

10.8 g Na2SO4/MM Na2SO4

Step2: Find out the moles of S:

moles S = moles Na2SO4

Step 3: Find out the mass of S.

mass S = moles S x MM S

Dimensional analysis:

? g S = 10.8 g Na2SO4 x (1mol Na2SO4/MM Na2SO4) x (1mol S/1mol Na2SO4) x (MM S/ 1 mol S) = (10.8/142.04)x32.1 = 2.44 g S

Answer:

Explanation:

Kandndkjwnww

Answer:

im only in 6th grade TwT

Explanation:

LOL

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.