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3 years ago

Which statement is true about a reaction represented by a chemical equation that shows energy as a reactant?

Chemistry

2 answers:

UNO [17]3 years ago

6 0

Answer:

A) It absorbs energy.

Explanation:

A>P>E>X

vovikov84 [41]3 years ago

5 0

A. It absorbs energy.
reactants are located on the left side of the equation, meaning energy among with other reactants were needed to get the reaction going, so it absorbed energy, which is also the endothermic process. The opposite of that would be having energy on the right side with the products which means that the reaction would've released energy which is the exothermic process. Hope this helps!

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Correct answer. 9. The mass number of an element is the number of?

Elis [28]

Answer:

The mass number is defined as the total number of protons and neutrons in an atom. The number of neutrons = mass number − atomic number.

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3 years ago

The density of a 20.3 m ch3oh (methanol) solution is 0.858 g/ml. what is the molality of this solution? h2o is the solvent.

aliina [53]

The difference of molarity and molality is in the units:

Molarity is expressed in mol solute / L solution;

<span> while Molality is expressed in mol solute / kg solvent</span>

Since no other data is given, let us assume that the contribution of methanol on the total volume is negligible. So that,

Molality = (20.3 mol / L) (1 L / 0.858 kg)

Molality = 23.66 m

<span>The closest answer is letter C. 23.7 m</span>

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4 years ago

"The occurrence of the all-or-none ________________ results from depolarization in response to rapid entry of ____________ ions

Leno4ka [110]

Answer:

The correct answer is

B.action potential; sodium

Explanation:

Action potential

Nerve action potentials are initiated and propagated in an all or nothing fashion. The all-or nothing action potential is initiated when the membrane potential threshold value is reached

Sodium

Voltage-gated Na⁺ ions membrane channels open enabling an influx of sodium ions to rapidly enter the cell, raising the membrane potential up to +40 mV

Depolarization in the neuron is based ion exchange channels opening to allow the rapid influx of sodium ions (Na⁺) and the efflux of potassium ions (K⁺)

The neuronal responds rapidly and for short periods to the activation of an ion channel receptor by a drug or natural neurotransmitter is rapid rel

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3 years ago

Read 2 more answers

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

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4 years ago

A man weighing 48 N climbs 82.0 meter ladder how much work is done

Karolina [17]

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3 years ago