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3 years ago

How many molecules are in 6.0g of sodium phosphate?

Chemistry

1 answer:

Phantasy [73]3 years ago

3 0

Answer:

2.2 x 10²² molecules.

Explanation:

Firstly, we need to calculate the no. of moles in (6.0 g) sodium phosphate:

no. of moles = mass/molar mass = (6.0 g)/(163.94 g/mol) = 0.0366 mol.

It is known that every mole of a molecule contains Avogadro's number (6.022 x 10²³) of molecules.

using cross multiplication:

1.0 mole of sodium phosphate contains → 6.022 x 10²³ molecules.

0.0366 mole of sodium phosphate contains → ??? molecules.

∴ The no. of molecules in 6.0 g of sodium phosphate = (6.022 x 10²³ molecules)(0.0366 mole)/(1.0 mole) = 2.2 x 10²² molecules.

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10 molecules SO2 to moles

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1.66 * 10^-23 moles

Explanation:

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4 years ago

What type of energy does a rock resting on top of the hill have?

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Read 2 more answers

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

4 years ago

What is the frequency and energy per quantum (in Joules) of :

viktelen [127]

(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

(b) f = 1.20 × 10¹⁰ Hz, E = 7.96 × 10⁻²⁴J.

<h3>Explanation</h3>

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at $3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1}$ in vacuum.

$f = \dfrac{c}{\lambda}$

where

f is the frequency of the electromagnetic radiation,
c is the speed of light, and
$\lambda$ is the wavelength of the radiation.

(a)

Convert all units to standard ones.

$\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}$ .

The unit of $f$ shall also be standard.

$f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}$ .

For each particle,

$E = h\cdot f$ ,

where

$E$ is the energy of the particle,
$h$ is the planck's constant where $h = 6.63\times 10^{-34}\;\text{J}\cdot\text{s}^{-1}$ , and
$f$ is the frequency of the particle.