A reaction between solid sodium hydroxide and gaseous carbon dioxide produced solid sodium carbonate and liquid water. Write out a balanced equation for this reaction.

Answer:

Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?

Answer: Tristan is most like the Golgi Body

Explanation:

The sample of white powder contains 47.1% K2CO3 and 0.39% Na2CO3.

Molar mass of sodium carbonate = 106 g/mol

Molar mass of potassium carbonate = 138 g/mol

Number of moles of HNO3 = 10/1000 L × 0.8903 M = 0.008903 moles

Mass of HNO3 =  0.008903 moles × 63 g/mol = 0.56 g

Mass of sample added = 1.091 g

Mass of sample left over = 0.573 g

Mass of sample reacted = 1.091 g - 0.573 g = 0.518 g

The reacted sample contains xg of Na2CO3 and (0.518 - x) g K2CO3.

Na2CO3 + 2HNO3 --> 2NaNO3 + CO2 + H2O

106g of Na2CO3 reacts with 126g of HNO3

x g of Na2CO3 reacts with (126 × x/106)g of HNO3

K2CO3 + 2HNO3 --> 2KNO3 + CO2 + H2O

138 g of K2CO3 reacts with 126 g of HNO3

(0.518 - x) g of K2CO3 reacts with [(0.518 - x) × 126/138] g

Total mass of HNO3 used;

1.19x + 0.47 + 0.91x = 0.56

2.1x + 0.47 = 0.56

2.1x = 0.56 - 0.47

2.1x = 0.09

x =  0.09/2.1

x = 0.0043 g

Mass of K2CO3 = (0.518 - x) g = 0.518 - 0.0043 = 0.5137 g

Mass percent of K2CO3  = 0.5137 g/ 1.091 g × 100/1 = 47.1%

Mass percent of Na2CO3 = 0.0043/1.091 g × 100/1 = 0.39%

Learn more: https://brainly.com/question/9743981

Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

Learn more in:

https://brainly.com/question/14020637

Answer:

D. works with bones to help you move

Explanation:

Your muscles and tendons are what help bones move.

Answer:

12 kilograms is larger.

Explanation:

Of the three units, the kilogram is the largest and the milligram is the smallest. The prefix “kilo” means a thousand and “milli” means one-thousandths. A gram is the basic unit of mass.

The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

We'll begin by calculating the number of half-lives that has elapsed.

Original amount (N₀) = 0.240 mole

Amount remaining (N) = 1.88×10¯³ mole

N = 1/2ⁿ × N₀

1.88×10¯³ = 1/2ⁿ × 0.240

Cross multiply

1.88×10¯³ × 2ⁿ = 0.240

Divide both side by 1.88×10¯³

2ⁿ = 0.240 / 1.88×10¯³

2ⁿ = 128

2ⁿ = 2⁷

Thus, 7 half-lives has elapsed

Number of half-lives (n) = 7

Half-life (t½) = 11.4 days

t = n × t½

t = 7 × 11.4

Therefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days

Learn more: https://brainly.com/question/13266270

Answer:

D

Explanation:

D Is The Answer Fella, My Head Hurt Really Bad

[tex]pH = -\log[H^{+}] \\\\\implies \log[H^{+}] = -pH\\\\\implies [H^{+}] = 10^{-pH}\\\\\implies [H^{+}] = 10^{-4.83} = 0.000015[/tex]

Answer:

qrtyuioplkjhgfdssssssazxcvbn

Answer:

44

Explanation:

Molar mass of Argon, Ar = 39.95 g Mass of 1 atom, m = 39.95 / (6.023 * 10^23) = 6.64 * 10^(-23) g =6.64 * 10^(-26) Kg Now, Kinetic Energy, KE = (1/2) * m * v^2 where velocity,v = 650 m/s Mass of 1 atom, m = 6.64 *...

- BRAINLIEST answerer

Answer:

1.40×10-20

Explanation:

Oh calculate the mass of 1 atom of argon the molar mass of argon 39.95g that is 6.0²×10²³atoms maybe...?

Answer: Atomic number

Explanation:

I hope this helps you!

Answer:

Explanation:

It''s chemical formula is Na3PO4

Answer:

c) single bonds, double bonds, carbon atoms and hydrogen atoms

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: https://brainly.com/question/12913997

Answer:

KNO2, KBr

Explanation:

Chemical compounds are any substance composed of identical molecules consisting of atoms of two or more chemical elements. So NO2 and KBr are compounds, Br2 and Fe are not.

Answer:

Answer D. Picture II shows a chemical change, because the same substance changes form

Explanation:

This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal

Answer:

D Im 90% Sure

Explanation:

If its not right i owe you one I did this one before

Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.

Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.

Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.

Learn more: https://brainly.com/question/967776

The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.

The correct answer is sp3.

The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.

The valence electrons are in p orbitals and p orbit mixed after the s orbitals.

Therefore, the correct answer is sp3

For more information, refer to the link:--

https://brainly.com/question/3523174

The heat produced by 4.5 moles of magnesium when burnt is 2700 kJ.

A thermochemical reaction is a reaction in which the amount of heat lost or gained is included in the reaction equation. The thermochemical reaction equation for the combustion of magnesium is shown as follows;

2 Mg(s) + O2(g) → 2 MgO(s) + 1200 kJ

From the reaction equation;

2 moles of magnesium produced 1200 kJ of heat

4.5 moles of magnesium will produce 4.5 moles × 1200 kJ/2 moles

= 2700 kJ

Learn more: https://brainly.com/question/14281129

Answer: 1.324L

Explanation: use Boyles law, sorry so late!

Answer:

go get the stuff.

Explanation:

Answer:

hydrogen bond is a type of dipole-dipole interaction; it is not a true chemical bond it is a mere electrostatic attraction. These attractions can occur between molecules or within different parts of a single molecule. Intramolecular hydrogen bonds are those which occur within one single molecule.

Explanation:

CAN I GET BRAINLIEST

This is the answer i hope it helps

A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?

Answer:

landslide mudflows slump and creep

Explanation:

no explanation

Answer:

[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

Answer:

D  BLESSS Y

Explanation:

Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.

If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"

Learn more: https://brainly.com/question/14281129

The temperature of an object will increase if the kinetic energy of the particle increases.

Kinetic energy is an energy that is said to be in motion. According to the kinetic molecular theory of ideal gas, the particles of the gas are usually moving in constant random motion and they exert no force on each other.

Also, the temperature varies directly proportional to the average kinetic energy of the gas particles. As a result, when the kinetic energy of the particles increases, the temperature will also increase.

Learn more about the kinetic theory of gas here:

https://brainly.com/question/9949658?referrer=searchResults

When the value of secondary quantum number (l) = 1, the magnetic quantum number can be from -1 to +1. The answer would be false.

The angular quantum number (l) can take any integer between 0 and the value of the principal quantum number (n) less by 1.

In other words, l can be from 0 to n-1.

The magnetic quantum number (m), on the other hand, can take any value between -l to +l.

For example, if l = 1, m can be -1,0, and +1.

More on quantum numbers can be found here: https://brainly.com/question/16977590

#SPj1