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
Over [174]
3 years ago
8

What is the job of a scientist?

Chemistry
2 answers:
12345 [234]3 years ago
7 0
B: to ask and answer scientific questions
Rudiy273 years ago
5 0

Answer:

B. To ask and answer scientific questions

You might be interested in
What is the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml?
Firdavs [7]

Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,  

M_1 = molarity of stock CuSO_4 solution = 2.5 M

V_1 = volume of stock CuSO_4solution = 5 ml  

M_1 = molarity of diluted CuSO_4 solution = ?

V_1 = volume of diluted CuSO_4 solution = 750 ml

Putting in the values we get:

2.5\times 5=M_2\times 750

M_2=0.017

Therefore the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

6 0
3 years ago
Write down the reaction of sodium hydioxide with<br>hydrochloric acid in words and symbols.​
ZanzabumX [31]
Hello!

the reaction:
sodium hydroxide + hydrochloric acid = sodium chloride

NaOH + HCl -> NaCl + H2O

if this helps, click “thanks” :)

- emily
5 0
3 years ago
Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=2:2=1:1

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=3:3=1:1

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


4 0
3 years ago
When the electron of a hydrogen atom moves into a higher energy orbit, what is the state of the atom?
faltersainse [42]
Answer is: <span>excited state.
In </span>excited state, hydrogen has<span> higher </span>energy<span> than in the </span>ground state (state with lowest energy). H<span>ydrogen atom has one </span>electron<span> in the lowest possible </span>orbit<span> (1s), when atom absorbs</span><span> energy</span><span>, the electron move into an excited state (quantum numbers greater than the minimum possible). </span>Electron lifetime in excited state is short.
4 0
3 years ago
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

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

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
Other questions:
  • I need help with this problem
    7·1 answer
  • Explain, in terms of intermolecular forces, why ammonia has a higher boiling point than the other compounds in the ta
    5·1 answer
  • A simple equation relates the standard free‑energy change, ΔG∘′, to the change in reduction potential. ΔE0′. ΔG∘′ = −nFΔE0′ The
    8·1 answer
  • Under the same conditions of temperature and pressure, which of the following gases would behave most like an ideal gas? A)He(g)
    14·1 answer
  • What would you expect to have the highest melting point?
    11·1 answer
  • Need help please on this EDPUZZLE
    6·1 answer
  • Match the standard reduction potentials for each half reaction and the electric potential for the voltaic cell.
    7·1 answer
  • 1.25 x 1025 molecules of glucose (C H202) is<br> how many moles of glucose?
    11·1 answer
  • 5. When sodium, a metal, is added to water, it reacts vigorously to form two new substances
    12·1 answer
  • Which type of region would be most likely to support chemical weathering?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!