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Kryger [21]
3 years ago
7

What is bromin what is bromine what is bromine

Chemistry
2 answers:
Vika [28.1K]3 years ago
5 0

Answer:

Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly coloured gas. Its properties are thus intermediate between those of chlorine and iodine

yulyashka [42]3 years ago
3 0

Answer:

Bromine is a chemical element with the symbol Br and atomic number 35. It is the third-lightest halogen, and is a fuming red-brown liquid at room temperature that evaporates readily to form a similarly colored gas.

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Choose all the correct information concerning chemical reactions: Involves small amounts of energy. Some of the mass is converte
suter [353]
<span>A chemical reactions :
Involves small amounts of energy.
</span><span> Only electrons are involved.
</span><span>The mass remains the same.</span>
8 0
3 years ago
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Identify a process that is NOT reversible. A. melting of steel B. freezing water C. melting of ice D. frying an egg E. depositio
vredina [299]
I’m pretty sure it would be D. Frying an egg
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3 years ago
A 20.0 milliliter sample of 0.200 molar K₂CO₃ solution is added to 30.0 milliliters of 0.400 molar Ba(NO₃)₂ solution. Barium car
sveticcg [70]

Answer:

0.32M

Explanation:

<u>Step 1:</u>  Balance the reaction

K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3

We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution

SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3

<u>Step 2: </u>Calculate concentration

To find the concentration of the barium cation we use the following equation:

Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>

<u />

<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3

[Ba2+] = 0.32 M

The concentration of Barium ion in solution is 0.32 M

6 0
3 years ago
Name the following aromatic hydrocarbon:
Tresset [83]

3-Methylpentane is the IUPAC name for the substance.

whether in a continuous chain or a ring, the longest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature. According to a precise set of priorities, all deviations—whether they involve numerous bonds or atoms other than carbon and hydrogen—are denoted by prefixes or suffixes.

+3-Methylpentane is the IUPAC name for the substance in question. It has a lengthy chain of 5 carbon atoms, which gives it the prefix pent-, and a single bond is what gives it the postfix -ane (alkane). Given that the methyl group is present at the third carbon, it is 3-methylpentane.

Learn more about IUPAC Nomenclature here-

brainly.com/question/14379357

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8 0
2 years ago
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

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

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

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

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

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3 years ago
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