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2 years ago

5

La Pokebola del gráfico realiza un MCU con una rapidez angular de 2π 9 rad s ⁄ . Determine el tiempo que emplea para ir desde A

hasta B

1 answer:

ddd [48]2 years ago

7 0

Fuvghbgbhj kkiggggbb ghhhb

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A common function of proteins and carbohydrates in the plasma membrane is...

Vika [28.1K]

I believe your answer is B.

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3 years ago

The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this

Semmy [17]

Answer:

The molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

Explanation:

Given that,

Mass of sample = 21.5 g

0.07 % (m/m) of copper (II) sulfate in plant fertilizer

This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present

So, in 20 g of plant fertilizer

, $=\frac{0.07}{100}\times 21.5}\\=0.0151g$ of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Mass of solute (copper (II) sulfate) = 0.0151 g

Molar mass of copper (II) sulfate = 159.6 g/mol

Volume of solution = 2.0 L

$\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M$

The chemical equation for the ionization of copper (II) sulfate follows:

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = $4.73\times 10^{-5}M$

Hence, the molar concentration of $Cu^{2+}$ ions in the given amount of sample is $4.73\times 10^{-5}M$

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3 years ago

Is it possible for a diprotic acid to have Ka1?

Masja [62]

Answer:

Yes, it is possible.

Explanation:

A diprotic acid is an acid that can release two protons. That's why it is called diprotic.

Monoprotic → Release one proton, for example Formic acid HCOOH

Triprotic → Releases three protons, for example H₃PO₄

Polyprotic → Release many protons, for example EDTA

it is a weak acid.

In the first equilibrum, it release proton, and the second is released in the second equilibrium. So the first equilibrium will have a Ka1

H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁

HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂

The HA⁻ will work as an amphoterous because, it can be a base or an acid, according to this:

HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂

HA⁻ + H₂O ⇄ OH⁻ + H₂A Kb₂

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3 years ago

Is bread a molecule or a compound?

Morgarella [4.7K]

it's molecule.......

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2 years ago

Read 2 more answers

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and

swat32

Answer:

$\%\ mass\ of\ CaCO_3=93.37\ \%$

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = 293.15 K

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.mmHg/K.mol

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K

⇒n of $CO_2$ produced = 0.0493 moles

According to the reaction:-

$CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2$

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of $CaCO_3$ = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0493\ mol= \frac{Mass}{100.0869\ g/mol}$

$Mass_{CaCO_3}=4.93\ g$

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

$\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100$

$\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100$

$\%\ mass\ of\ CaCO_3=93.37\ \%$

3 0

3 years ago