How many moles are in 5.6 grams of Na?

Chemistry

2 answers:

a_sh-v [17]3 years ago

7 0

There are 0.244 moles in 5.6 grams of Na

Lubov Fominskaja [6]3 years ago

4 0

Explanation:

n=m/M

M(Na)=23

m(Na)=5.6

n=5.6/23

n=0.243moles

You might be interested in

How can valence electrons be used to predict the type of compounds formed?

Alex17521 [72]

Valence electrons are the electrons in the outermost shell. Those are the electrons on an atom that can be gained or lost in a chemical reaction.
Elements that are left on the periodic table have relatively few valence electrons, and can form ions more easily by losing their valence electrons to form positively charged cations.
Nonmetals are further to the right on the periodic table, so they gain electrons relatively easily and lose them with difficulty.

7 0

3 years ago

Complete the equation for the reaction of sulfuric acid and sugar (sucrose). C12H22O11 + 11 H2SO4 ? 12 + 11 H2SO4 + 11

serg [7]

Answer :

The complete equation for the reaction of sulfuric acid and sugar is,

$C_{12}H_{22}O_{11}+11 H_2SO_4\rightarrow 12C+11H_2SO_4+11H_2O$

By the stoichiometry of the reaction,

1 mole of sucrose react with the 11 moles of sulfuric acid to give 12 moles of carbon and 11 moles of water.

In this reaction, sulfuric acid react with sucrose (sugar). It dehydrates the sugar molecules which means it eliminates the water.

7 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$