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Agata [3.3K]
3 years ago
8

Who was the first person known to have used the terms genus and species when classifying organisms?

Chemistry
1 answer:
velikii [3]3 years ago
4 0

C Linnaeus was the first person known to have used the terms genus and species when classifying organisms.

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Which ions remain in solution after pbi2 precipitation is complete? express your answers as ions separated by a comma?
olga nikolaevna [1]

Answer;

K+ and NO3- ions

Explanation;

The main ions remaining are K+ and NO3- ions after pbi2 precipitation is complete.

However; There will always be tiny amounts of Pb2+ and I- ions, but most of them are in the solid precipitate.

7 0
4 years ago
What are some<br> double replacement reactions you see every day
Cerrena [4.2K]

Answer:

  • Sodium Bicarbonate (β-3) + Vinegar
  • Lead Nitrate + Potassium iodide

Explanation:

Baking Soda and vinegar cause an explosion, in which the bicarbonate and vinegar are replaced by nitrate (∨) and oxide (Ф.) When you combine lead nitrate (Δω) with potassium iron, you also see the ingredients you combined disappear, which shall cause a replace reaction

6 0
3 years ago
There is a gas at 780 mm of Hg, in a volume of 5 liters and a temperature of 37 ​ C, the volume is changed to 5.5 liters and the
HACTEHA [7]

Answer:

32.8 C

Explanation:

- Use combined gas law formula and rearrange.

- Hope that helped! Please let me know if you need further explanation.

6 0
4 years ago
Số hợp chất tố đa của nguyên tố này với nguyên tố khác theo hóa trị của nó
SVETLANKA909090 [29]

Answer:

Carbon

Explanation:

Carbon

5 0
3 years ago
What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
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