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3 years ago

Formula for the compound Tetraoxygen nonabromide?

Chemistry

1 answer:

Vikentia [17]3 years ago

3 0

Answer:

O4Br9

Explanation:

Tetra means 4 and nona means 9. Just use these prefixes to decide how many of each element you have.

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Given the balanced equation representing a reaction:

dimaraw [331]

The answer is (4) synthesis. Synthesis reaction means that two or more reactants combine directly to one production. Substitution or single replacement means that one element of a compound is replaced by another element. Double replacement means that two ionic reactants exchange ions to form two new productions.

3 0

3 years ago

a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element

Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0

3 years ago

HELP PLEASE<br> Chemical name for As2S3

irina1246 [14]

Answer:

Chemical name for As2S3 is Arsenic sulfide

Explanation:

4 0

3 years ago

Read 2 more answers

Draw the most stable lewis structure of acrolein.

NeX [460]

See image below for the lewis structure of acrolein

3 0

4 years ago

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Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

2. Volume from density and mass

$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

3. Mass from density and volume

$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water =<u> 12.8 mL</u>

Volume of object = 11.8 mL

$\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}$

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.

6 0

3 years ago