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kipiarov [429]
3 years ago
11

How can a periodic table help a chemist write the chemical name of a compound from its chemical formula?

Chemistry
2 answers:
madam [21]3 years ago
7 0
It's B. by identifying the elements. 

Reasons: 
prefixes are only used for covalent compounds. 

You don't include the number of atoms when you write the chemical name of a compound. 

Subscripts are basically numbers that tell how many atoms of each element is in a compound. It does not identify the name of an element. 



creativ13 [48]3 years ago
4 0

Answer:

B) by identifying the elements

Explanation:

EDGE 2020 brainliest please

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The atmospheric pressure in Francisco on ascertain day was 97.6kpa what was the pressure on mmhg
vodka [1.7K]

Answer:

732.0601 mmHg

Explanation:

Given data:

Pressure = 97.6 KPa

Given pressure in mmHg = ?

Solution:

Kilo pascal and millimeter mercury both are units of pressure.

Kilo pascal is denoted as "KPa"

Millimeter mercury is denoted as " mmHg"

Kilo pascal is measure of force per unit area while also define as newton per meter square.

It is manometric unit of pressure. It is the pressure generated by column of mercury one millimeter high.

Conversation of kilopascal to mmHg:

97.6 × 7.501 = 732.0601 mmHg

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What evidence in Side View lets you know that upwelling is happening in these locations?
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Answer:

Image result for What evidence in Side View lets you know that upwelling is happening in these locations?

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Explanation:

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3 years ago
A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.
Alja [10]

Answer:

  • <u>Question 1: 0.2J/(gºC)</u>
  • <u>Question 2: 6,000J</u>
  • <u>Question 3: 300J</u>
  • <u>Question 4: 80g</u>
  • <u>Question 5: 74ºC</u>
  • <u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

  • Q = m × C × ΔT

<em />

Substitute and solve for C:

  • 566J = 20g × C × (195ºC - 35ºC)
  • C = 566J / (20g × 160ºC)
  • C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

  • Q = m × C × ΔT

<em />

Substitute and compute:

  • Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)
  • Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

  • Q = m × C × ΔT
  • Q = 50g × 0.420J/gºC × (35ºC - 50ºC)
  • Q = 315J

Round to one significant figure (because 50g has one significant figure)

  • Q = 300J

<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

  • Q = m × C × ΔT

Solve for m:

  • m = Q / (C × ΔT)

Substitute and compute:

  • m = 3,000J / (0.712J/gºC × 50ºC)
  • m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. <em>If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.</em>

<em />

  • Q = m × C × ΔT

<em />

Q is negative, since it is released.

Substitute and solve for T:

  • - 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

  • T = - 7500J / 400g × 4.184J/gºC) + 70ºC

  • T = 74ºC

<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

<em />

Question 6. <em>How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC</em>

  • Q = m × C × ΔT

Subsitute:

  • 9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

  • m = 9,500J / (4.184J/gºC × 50ºC)

  • m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

  • m = 50g.
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