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4 years ago

6.33 x 1024 atoms of C are how many moles?

Chemistry

1 answer:

Readme [11.4K]4 years ago

7 0

Answer:

Explanation:

just use avogardo's number which show that 1 mole has 6.022×10^23 atoms. so if you have 6.33 x 10^24 atoms of C, just divide that by 6.022x 10^23 and plug it into a calculator.

(6.33x10^24) / (6.022 x 10^23)

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Translate the skeleton chemical equation:

Minchanka [31]

Answer:

MgBr₂ + AgNO₃ => Mg(NO₃)₂ + AgBr

Explanation:

Find the element symbol and charge of each element on the periodic table. For polyatomic ions (nitrate), reference your polyatomic ions chart. Use the "partner's charge" rule to find the number of atoms in each compound.

Charges are written as superscripts. "1" is usually not written, just the + or - sign. The charge of silver is 1, which is the (I) bracket roman numeral 1. It is indicated like that because it is multivalent, meaning it has more than one possible charge.

<u>Write each element as an ion</u> (with the charge).

Magnesium is Mg²⁺

Bromide is Br⁻

Silver(I) is Ag⁺

Nitrate is (NO₃)⁻

<u>Write each compound.</u>

REACTANTS SIDE

Magnesium bromide

Mg²⁺Br⁻ Cross over the partner's charge. Since Br is charge 1, Mg has 1 atom. Since Mg has charge 2, Br has 2 atoms.

MgBr₂

Silver(I) nitrate

Ag⁺(NO₃)⁻

AgNO₃ Both have 1 atom because each partner's charge was 1. You do not need to write brackets if nitrate only has 1 atom.

PRODUCTS SIDE

Magnesium nitrate

Mg²⁺(NO₃)⁻

Mg(NO₃)₂ Nitrate has 2 atoms because magnesium's charge is 2.

Silver(I) bromide

Ag⁺Br⁻

AgBr Both have 1 atom.

Write the compounds into an equation. Reactants go on the left side, products go on the right side. Between the reactants and products, write an arrow.

MgBr₂ + AgNO₃ => Mg(NO₃)₂ + AgBr

6 0

4 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

4 years ago

Give the chemical symbol for the element with the ground‑state electron configuration [ Ar ] 4 s 2 3 d 1 . symbol: Determine the

Rus_ich [418]

Answer:

Sc (Scandium) has the given electronic configuration.

Explanation:

The given electronic configuration is [Ar] $4s^{2}3d^{1}$ .

The last electron enters the d-subshell and hence is a d-block element known as Scandium with chemical symbol Sc.

For 4s subshell

n=4,l=0 and m ranges from -l to +l so m=0.

For 3d subshell

n=3,l=2 and m ranges from -l to +l so m can take values -2,-1,0,+1,+2

Note:

l values for subshells:

s : 0

p : 1

d : 2

f : 3 and so on.

5 0

3 years ago

Water enters a 4.00-m3 tank at a rate of 6.33 kg/s and is withdrawn at a rate of 3.25 kg/s. The tank is initially half full.

olga55 [171]

Answer:

At the start of the process, the volume not occupied by the water is 2 m3

Explanation:

At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).

Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).

The volume in time will be

$V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]$

8 0

3 years ago

A sample of fluorine gas occupies 810 milliliters at 270 K and 1.00 atm. What volume does the gas occupy when the pressure is do

sertanlavr [38]

Answer:

The answer to your question is Volume = 600 ml

Explanation:

Data

Volume 1 = 810 ml

Temperature 1 = 270°K

Pressure 1 = 1 atm

Volume 2 = ?

Pressure = 2 atm

Temperature 2 = 400°K

Process

1.- To solve this problem, use the Combine gas law.

P1V1 / T1 = P2V2 / T2

- Solve for V2

V2 = P1V1T2 / T1P2

2.- Substitution

V2 = (1)(810)(400) / (270)(2)

3.- Simplification

V2 = 324000 / 540

4.- Result

V2 = 600 ml

8 0

3 years ago

Read 2 more answers