5. What is the identity of the element with the electron configuration 1s22s22p 3s 3p64523d 10?

One half of a balanced chemical equation is shown. 3Mg(OH)2 + 2H3PO4

nasty-shy [4]

Answer:

3Mg(OH)2 + 2H3PO4 = Mg3(PO4)2 + 2H3O

Explanation:

8 0

4 years ago

Giải thích tại sao chén sấy (hoặc chén nung) dùng chứa tủa phải xử lý cùng với điều kiện xử lý tủa? Thời gian sấy (hoặc nung) bắ

nevsk [136]

Hi how are you doing today Jasmine Jasmine

3 0

3 years ago

A certain compound of bromine and fluorine is used to make UF6, which is an important chemical in the processing and reprocessin

Aleksandr [31]

Answer: The empirical formula is $BrF_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Br= 58.37 g

Mass of F = (100-58.37) = 41.63 g

Step 1 : convert given masses into moles.

Moles of Br= $\frac{\text{ given mass of Br}}{\text{ molar mass of Br}}= \frac{58.37g}{80g/mole}=0.73moles$

Moles of F = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{41.63g}{19g/mole}=2.2moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Br = $\frac{0.73}{0.73}=1$

For F = $\frac{2.2}{0.7}=3$

The ratio of Br: F= 1 : 3

Hence the empirical formula is $BrF_3$

8 0

3 years ago

Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine

Yuliya22 [10]

Answer:

194 g/mol

Explanation:

1) Content of C:

All the C atoms in the 1.000 mg of caffeine will be found in the 1.813 mg of CO₂.

Mass of C in 1.813 mg of CO₂

Since the atomic mass of C is 12.01 g/mol and the molar of of CO₂ is 44.01, there are 12 mg of C in 44 mg of CO₂ and you can set the proporton:

12.01 mg C / 44.01 mg CO₂ = x / 1.813 g CO₂

⇒ x = 1.813 × 12.01 / 44.01 g of C = 0.49475 mg of C

Number of moles of C

number of moles = mass in g / atomic mass = 0.49475×10⁻³ g / 12.01 g/mol = 4.1195×10⁻⁵ moles = 0.041195 milimol

2) Content of H

All the H atoms in the 1.000 mg of caffeine will be found in the 0.4639 mg of H₂O

Mass of H in 0.4639 mg of H₂O

Since the atomic mass of H is 1.008 g/mol and the molar of of H₂O is 18.015 g/mol, there are 2×1.008 mg of H in 18.015 mg of H₂O and you can set the proporton:

2×1.008 mg H / 18.015 mg H₂O = x / 0.4639 mg H₂O

⇒ x = 0.4639 mg H₂O × 2 × 1.008 mg H / 18.015 mg H₂O = 0.051913 mg H

Number of moles of H

number of moles = mass in g / atomic mass = 0.051913 ×10⁻³ g / 1.008 g/mol = 5.1501× 10⁻⁵ moles = 0.051501 milimol

3) Content of N

All the N atoms in the 1.000 mg of caffeine will be found in the 0.2885 mg of N₂

Mass of N in 0.2885 mg of N₂ is 0.2885 mg

Number of moles of N

number of moles = mass in g / atomic mass = 0.2885 ×10⁻³ g / 14.007 g/mol = 2.0597× 10⁻⁵ moles = 0.020597 milimol

4) Content of O

The mass of O is calculated by difference:

Mass of O = mass of sample - mass of C - mass of H - mass of N

Mass of O = 1.000 mg - 0.49475 mg C - 0.051913 mg H - 0.2885 mg N

Mass of O = 0.1648 mg

Moles of O = 0.1648 × 10 ⁻³ g / 15.999 g/mol = 1.0303×10⁻⁵ mol = 0.01030 milimol

5) Ratios

Divide every number of mililmoles by the smallest number of milimoles:

C: 0.041195 / 0.01030 = 4
H: 0.051501 / 0.01030 = 5
N: 0.020597 / 0.01030 = 2

O: 0.01030 / 0.01030 = 1

C: 4
H: 5
N: 2
O: 1

6) Empirical formula:

C₄H₅N₂O₁

7) Calculate the approximate mass of the empirical formula:

4 × 12 + 5 × 1 + 2 × 14 + 1 × 16 = 97 g/mol

So, since that number is not between 150 and 200 g/mol, multiply by 2: 97 × 2 = 194, which is between 150 and 200.

Thus, the estimate is 194 g/mol

7 0

3 years ago

isotopes of an element will always have the same a. number of protons. b. number of neutrons. c. number of electrons. d. mass.

Nastasia [14]

Your answer is same number of protons

3 0

3 years ago

5. What is the identity of the element with the electron configuration 1s22s22p 3s 3p64523d 10?​

5. What is the identity of the element with the electron configuration 1s22s22p 3s 3p64523d 10?