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AlekseyPX
3 years ago
11

Which tags do not depend on a silicon microchip and use plastic or conductive polymers instead of silicon-based microchips allow

ing them to be washed or exposed to water without damaging the chip?
Chemistry
1 answer:
user100 [1]3 years ago
6 0

Answer: Chipless RFID tags

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Convert 15.2 moles of K to atoms of K.
choli [55]

Answer:

\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

Multiply by the given number of moles: 15.2

15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

The moles of potassium cancel.

15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }

The denominator of 1 can be ignored.

15.2 * {6.022*10^{23} \ atoms \ K }{

Multiply.

9.15344*10^{24} \ atoms \ K

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

9.15*10^{24} \ atoms \ K

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

3 0
2 years ago
Do you think one mole of the different substances has the same amount?​
vagabundo [1.1K]

It's because of Avagadro's number.

  • It is known as Avagadro's constant too.
  • It states that at constant temperature one mole of any substance contain same number of atoms i.e n no of atoms

where

\\ \sf\longmapsto n=6.022\times 10^{23}

5 0
2 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
2 years ago
I need help with this question please and thank you
Tcecarenko [31]
The answer should be B
6 0
3 years ago
The nucleus of unstable _____ of an element will decay leading to emission of radiation.question 1 options:moleculescationsanion
Marta_Voda [28]

Isotopes of same element has different number of neutrons with different masses and having same number of protons and electrons.

Radioactive isotopes are those isotopes which are radioactive in nature. The unstable nucleus results in the radioactivity process and this process will go on until the stable isotope (element) forms.

Thus, the nucleus of unstable isotopes of an element will decay leading to emission of radiation.

6 0
3 years ago
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