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skad [1K]
2 years ago
9

Question 8 Lithium is a solid phase of lithium still unknown to science. The only difference between it and ordinary lithium is

that Lithium forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Lithium . Round your answer to significant digit
Chemistry
1 answer:
trapecia [35]2 years ago
7 0

Answer:

the density of lithium will be equal to: 0.582 g/cm^3

Explanation:

We have the following data:

a = 430 pm = 4.3x10^-8 cm

molar mass of lithium MM = 6.97 g/mol

Since the cubic cell of lithium is cubic centered in the face, its value of Z will be equal to 4

Avogadro´s number AN = 6.022x10^23 particles/mol

using the following formula we will calculate the density of lithium:

d = (Z*MM)/(a^3*AN) = (4*6.97)/((4.3x10^-8)^3*(6.022x10^23)) = 0.582 g/cm^3

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If 7.36 g of sodium metal react, how many grams of sodium hydroxide can be produced
8090 [49]
You would also need to know the mol of sodium. After knowing the mol, write down the reaction;
Na + H20 -> NaOH + H2
Use the mol ratio, to find the amount of mol of sodium hydroxide. (Remember to balance the equation first)
After you have the necessary variables, use the following formula:
mass of sodium hydroxide = mol of sodium hydroxide x molar mass (relative/formula mass of sodium hydroxide)
That is how you will find the mass of sodium hydroxide.
8 0
3 years ago
Calculate the molarity of an hcl solution if a 10.00 ml sample requires 25.24 ml of a 1.600 m naoh solution to be neutralized.
AlladinOne [14]

Answer:

The molarity of the HCl solution should be 4.04 M

Explanation:

<u>Step 1:</u> Data given

volume of HCl solution = 10.00 mL = 0.01 L

volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L

<u>Step 2:</u> The balanced equation

HCl + NaOH → NaCL + H2O

Step 3: Calculate molarity of HCl

n1*C1*V1 = n2*C2*V2

Since the mole ratio for HCl and NaOH is 1:1 we can just write:

C1*V1 =C2*V2

⇒ with C1 : the molarity of HCl = TO BE DETERMINED

⇒ with V1 = the volume og HCl = 10 mL = 0.01 L

⇒ with C2 = The molarity of NaOH = 1.6 M

⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L

C1 * 0.01 = 1.6 * 0.02524

C1 = (1.6*0.02524)/0.01

C1 = 4.04M

The molarity of the HCl solution should be 4.04 M

6 0
3 years ago
Amestec 140 g solutie 9% cu 260 g solutie 12%, cu 60 g apa si x g solutie 4%. Ce c procentuala are noua solutie %?
Levart [38]

Answer:

PLEASE SOMEONE SOLVE THE QUESTION OR I'M GONNA FAIL

Explanation:

Purpose and Theory: A tablet of vitamin C contains the active ingredient called ascorbic acid (HC,H,0.).

Ascorbic acid is an organic acid which is a white power and is soluble in water.

In the following microtitration lab you will be grinding up a vitamin C tablet

, and using a 0.10 g sample of the

vitamin C powder, which will be dissolved in 20-40 mL of water. This solution will be used to determine the

quantity of ascorbic acid in a commercial vitamin C tablet.

Groups

Lab 2

# of Drops 1 or Drops 2 of Drops 3

96

91

94

Hadia

Data & Analysis:

Complete the missing information in the following data table [24]

1

2

3

Trial

Volume of NaOH (drops)

0 05mL drop

(NaOH (O 0050mol/L)

Volume of NaOH (ML)

Average Volume of NaOH (ml)

T I

O 0050

00050

00050

The balanced chemical reaction between ascorbic acid and sodium hydroxide is:

HC,H,O + NaOH NaC,H,Oye H,O,

1. Stoichiometrically calculate the mass (in grams) of ascorbic acid in the powdered sample. 

3 0
3 years ago
Classify each element as a metal, nonmetal, or semimetal. br in cr ca drag the appropriate items to their respective bins. view
goldenfox [79]
Below is the distribution of given elements and additional elements in their respective bins.

Iron, Magnesium, Aluminium, Calcium, Chromium and Indium are classified as Metals due to their hardness, conduction of current and heat, Solid state, formation of metallic bonding and prefers to form cations e.t.c

Oxygen, Neon and Bromine are classified as non-metals. Non-metals are mostly gases except for Bromine. They are highly volatile and prefers to form anions.

Semimetals are those metals which have both properties of metals and non-metals and are also called as metalloids.

6 0
3 years ago
Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0
1 year ago
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