1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
jek_recluse [69]
3 years ago
12

Is gold a compound or an element

Chemistry
1 answer:
timurjin [86]3 years ago
3 0
Gold is an element (Au) you can identify it on a periodic table
You might be interested in
1.) What are the two broad categories that elements can be divided into?
hoa [83]

Answer:

1) Metals and nonmentals

2) Elements: Oxygen (O_{2} \\) , Nitrogen (N_{2}

3) Compounds: Carbon Dioxide CO_{2}, Methane CH_{4}, Nitrogen DioxideNO_{2}

Explanation:

5 0
3 years ago
How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


8 0
3 years ago
What 100 plus 124000
kodGreya [7K]
100 + 124000 = 124100
6 0
3 years ago
Read 2 more answers
The density of water is 1.00 gram/milliliter. What is the volume in milliliters of 1.00 mole of water? Express your answer to th
Kay [80]

Answer: 18.0152 milliliters

Explanation:

Hi, to answer this question we have to apply the next formula:

Water volume = water mass / water density

Since 1 mol of water weights 18.0152 grams

Replacing with the values and solving:

Water volume = 18.0152 g / 1 g /ml = 18.0152 milliliters

Feel free to ask for more if needed or if you did not understand something.  

5 0
4 years ago
If a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay, what will be the new element?
Charra [1.4K]

Answer:

The new element will be thorium-226 (²²⁶Th).

Explanation:

The beta decay is given by:

^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + \beta^{-} + \bar{\nu_{e}}

Where:

A: is the mass number

Z: is the number of protons  

β⁻: is a beta particle = electron

\bar{\nu_{e}}: is an antineutrino

The neutral atom has 88 electrons, so:

e^{-} = 88 = Z

Hence the element is radium (Ra), it has A = 226.

If Ra undergoes 2 rounds of beta minus decay, we have:

^{226}_{88}Ra \rightarrow ^{226}_{89}Ac + \beta^{-} + \bar{\nu_{e}}    

^{226}_{89}Ac \rightarrow ^{226}_{90}Th + \beta^{-} + \bar{\nu_{e}}

Therefore, if a neutral atom with 88 electrons undergoes 2 rounds of beta minus decay the new element will be thorium-226 (²²⁶Th).

I hope it helps you!    

8 0
3 years ago
Other questions:
  • A compound has a chemical composition of 97.6 g carbon, 4.9 g hydrogen, 52 g oxygen, 45.5 g nitrogen. What is the empirical form
    15·1 answer
  • Change 23 degrees Celsius to F
    11·2 answers
  • Hydrochloric acid and sodium hydroxide will produce which two compounds
    8·1 answer
  • A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
    15·1 answer
  • Is air a mixture or pure substance ​
    12·1 answer
  • A sample of an unknown substance has a mass of 0.465 kg. If 3,000.0 J of heat is required to heat the substance from 50.0°C to 1
    12·2 answers
  • Which type of orbital is shown?
    14·1 answer
  • Which of the following equations is correctly balanced? UO2(s)+4HF(l)→UF4(s)+H2O(l) NCl3(g)+3H2O(l)→NH3(g)+HClO(aq) 4NH3(g)+3O2(
    8·1 answer
  • Write the formula for each of the following...I'm not telling you what kind of compound it
    15·1 answer
  • What is pseudo - order reaction​
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!