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V125BC [204]
3 years ago
11

what is the momentum of a baseball with a mass of 2kg being thrown at a velocity of 85m/s forwards the hitter?​

Chemistry
1 answer:
Nutka1998 [239]3 years ago
3 0

Answer:

<h2>170 kg m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 2 × 85

We have the final answer as

<h3>170 kg m/s</h3>

Hope this helps you

You might be interested in
The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van
Goshia [24]

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: ΔT_{b}=i K_{b}.m, where ΔT_{b}= elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

K_{b}= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i=\frac{2.5}{0.512 X 1.83}

i=2.66= 2.7 (approx.)


6 0
3 years ago
What is the relative molecular mass (mr) of ammonia, formula NH3?
egoroff_w [7]

Answer:

APROXIMENTLY 17

Explanation:

lol i remebered from science class

7 0
3 years ago
What happens when sodium and sulfur combine
Eduardwww [97]

Answer:

Sodium sulfide is the chemical compound with the formula Na2S, or more commonly its hydrate Na2S·9H2O. Both the anhydrous and the hydrated salts are colorless solids. They are water-soluble, giving strongly alkaline solutions. When exposed to moist air, Na2S and its hydrates emit hydrogen sulfide, which smells like rotten eggs. Some commercial samples are specified as Na2S·xH2O, where a weight percentage of Na2S is specified. Commonly available grades have around 60% Na2S by weight, which means that x is around 3. Such technical grades of sodium sulfide have a yellow appearance owing to the presence of polysulfides. These grades of sodium sulfide are marketed as 'sodium sulfide flakes'.

Contents

1 Structure

2 Production

3 Reactions with inorganic reagents

4 Uses

4.1 Reagent in organic chemistry

5 Safety

6 References

Structure

Na2S adopts the antifluorite structure,[2][3] which means that the Na+ centers occupy sites of the fluoride in the CaF2 framework, and the larger S2− occupy the sites for Ca2+.

Production

Industrially Na2S is produced by carbothermic reduction of sodium sulfate often using coal:[4]

Na2SO4 + 2 C → Na2S + 2 CO2

In the laboratory, the salt can be prepared by reduction of sulfur with sodium in anhydrous ammonia, or by sodium in dry THF with a catalytic amount of naphthalene (forming sodium naphthalenide):[5]

2 Na + S → Na2S

Reactions with inorganic reagents

The sulfide ion in sulfide salts such as sodium sulfide can incorporate a proton into the salt by protonation:

S2−

+  H+ → SH−

Because of this capture of the proton ( H+), sodium sulfide has basic character. Sodium sulfide is strongly basic, able to absorb two protons. Its conjugate acid is sodium hydrosulfide (SH−

). An aqueous solution contains a significant portion of sulfide ions that are singly protonated.

S2−

+ H

2O {\displaystyle {\ce {<=>>}}}{\displaystyle {\ce {<=>>}}} SH−

+  OH−

 

 

 

 

(1)

SH−

+ H

2O {\displaystyle {\ce {<<=>}}}{\displaystyle {\ce {<<=>}}} H

2S +  OH−

 

 

 

 

(2)

Sodium sulfide is unstable in the presence of water due to the gradual loss of hydrogen sulfide into the atmosphere.

When heated with oxygen and carbon dioxide, sodium sulfide can oxidize to sodium carbonate and sulfur dioxide:

2 Na2S + 3 O2 + 2 CO

2 → 2 Na2CO3 + 2 SO2

Oxidation with hydrogen peroxide gives sodium sulfate:[6]

Na2S + 4 H2O2 → 4 H

2O + Na2SO4

Upon treatment with sulfur, polysulfides are formed:

2 Na2S + S8 → 2 Na2S5

Uses

Sodium sulfide is primarily used in the kraft process in the pulp and paper industry.

It is used in water treatment as an oxygen scavenger agent and also as a metals precipitant; in chemical photography for toning black and white photographs; in the textile industry as a bleaching agent, for desulfurising and as a dechlorinating agent; and in the leather trade for the sulfitisation of tanning extracts. It is used in chemical manufacturing as a sulfonation and sulfomethylation agent. It is used in the production of rubber chemicals, sulfur dyes and other chemical compounds. It is used in other applications including ore flotation, oil recovery, making dyes, and detergent. It is also used during leather processing, as an unhairing agent in the liming operation.

Reagent in organic chemistry

Alkylation of sodium sulfide give thioethers:

Na2S + 2 RX → R2S + 2 NaX

Even aryl halides participate in this reaction.[7] By a broadly similar process sodium sulfide can react with alkenes in the thiol-ene reaction to give thioethers. Sodium sulfide can be used as nucleophile in Sandmeyer type reactions.[8] Sodium sulfide reduces1,3-dinitrobenzene derivatives to the 3-nitroanilines.[9] Aqueous solution of sodium sulfide can be refluxed with nitro carrying azo dyes dissolved in dioxane and ethanol to selectively reduce the nitro groups to amine; while other reducible groups, e.g. azo group, remain intact.[10] Sulfide has also been employed in photocatalytic applications.[11]

Explanation:there you go

7 0
3 years ago
How much iron is present in 7.59 g of iron(iii) oxide? answer in units of g?
Verdich [7]
You need the unit of g however you must convert to moles before you can use the mole ratio to find the moles of iron, you used the molar mass of iron to find the grams of iron. since F e 2 o 3 

4 0
3 years ago
What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
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