Someone help What are the products in a chemical reaction?

Chemistry

2 answers:

Alisiya [41]3 years ago

7 0

Answer:

The substances that result from a chemical reaction.

Explanation:

The substances produced at the end of the reaction are known as the products.

Keith_Richards [23]3 years ago

3 0

Answer:

D) The substances that result in a chemical reaction.

In a chemical reaction the substances which takes part in it is called reactant and the result obtained is called product.

Catalyst are the substances which helps in increasing the chemical reaction

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mario62 [17]

Answer:

HI is least polar

Explanation:

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3 years ago

Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re

lapo4ka [179]

a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K

when ΔG° is Gibbs free energy

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 = 298 K

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

= -7739 J = -7.7 KJ

b) then, to get E° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

= - 2.7 x 10^-2 V

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3 years ago

Help, I need this quick. Someone please.

timofeeve [1]

Answer:

the answer to that question is d

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A hot air balloon of 257m^3 is made from 8.50kg of materials. The outside atmosphere is 1.00 atm, and 18C and pure N2 atmosphere

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3 years ago

Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in

avanturin [10]

5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate

5 0

3 years ago