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cestrela7 [59]
2 years ago
10

Pleas help with 2 and 4 for brainliest

Chemistry
1 answer:
Snezhnost [94]2 years ago
4 0

mass of pentane : = 30.303 g

moles of Al₂(CO₃)₃ : = 0.147

<h3>Further explanation</h3>

Given

1. Reaction

C₅H₁₂+8O₂→6H₂O+5CO₂.

45.3 g water

2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂

37.2 MgCO₃

Required

mass of pentane

moles of Al₂(CO₃)₃

Solution

1. mol water = 45.3 : 18 g/mol = 2.52

From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :

= 1/6 x mol H₂O

= 1/6 x 2.52

= 0.42

Mass pentane :

= mol x MW

= 0.42 x 72.15 g/mol

= 30.303 g

2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44

mol Al₂(CO₃)₃ :

= 1/3 x mol MgCO₃

= 1/3 x 0.44

= 0.147

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Determine the location of the last significant place value by placing a bar over the digit.
Vlad1618 [11]

Answer:

8,040  

0.0300  

699.5  

2.000 x 102

0.90100  

90, 100  

4.7 x 10-8  

10,800,000.0  

3.01 x 1021

0.000410

Explanation:

First remember the following rules of determining the last significant place value :

1. The digits from 1-9 are all significant and zeros between significant digits are also significant.

2.  The trailing or ending zeroes are significant only in case of a decimal number otherwise they are ignored. However starting zeroes of such a number are not significant.

Now observing above rules, lets determine the location of the last significant place value of each given example. I am determining the location by turning the last significant place to bold.

1) 8,040

8,040

Location of the last significant place value is 3 and bar is over last significant digit that is 4. Number is not decimal so ending zero is ignored. Every non zero digit is a significant.

2)  0.0300

0.0300

Location is 3 and bar is over 0. Number has a decimal point so ending zero is not ignored but starting zeroes are ignored.

3) 699.5

699.5

Location is 4 and bar is over 5.

4) 2.000 x 10²

2.000 x 10²

Location is 4 and bar is over 0. This is because the number is decimal so trailing zeroes cannot be ignored. Also if we convert this number it becomes:

200.0 so last significant digit is 0 and location of last significant digit is 4.

5) 0.90100

0.90100

Location is 5 and bar is over 0. This is because in a number with decimal point starting zeroes are ignored but trailing zeroes after decimal point are not ignored. So we count from 9 and last significant digit is 0.

6) 90, 100

90, 100

Location is 3 and bar is over 1. This is because it is not a number with decimal point. So the trailing zeroes are ignored. The count starts from 9 and last significant is 1.

7) 4.7 x 10⁻⁸

4.7 x 10⁻⁸

Location is 2 and bar is over 7. This is because the starting zeroes in a number with a decimal point are ignored. So the first digit considered is 4 and last significant digit is 7. If we expand this number:

4.7 x 10⁻⁸ =    0.000000047 = 0.000000047

Here the starting zeroes are ignored because there is a decimal point in the number.

8) 10,800,000.0

10,800,000.0

Location is 9 and bar is over 0.  Number has a decimal point so ending zero is not ignored and last significant figure is 0.

However if the number is like:

10,800,000. Then location would be 8 and bar is over 0.

9) 3.01 x 10²¹

3.01 x 10²¹

Location is 3 and bar is over 1. Lets expand this number first

3.01 x 10²¹ = 3.01 x 1000000000000000000000

                  = 3010000000000000000000

So this is the number:

3010000000000000000000

Since this is number does not have a decimal point so the trailing zeroes are ignored. Hence the count starts from 3 and the last significant figure is 1

10) 0.000410

0.000410

Location is 3 and bar is over 0. This is because the number has a decimal point so the ending zero is not ignored but the starting zeroes are ignored according to the rules given above. Hence the first significant figure is 4 and last significant figure is 0.

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3 years ago
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If 65,000 J of energy were transferred to 450 g of water at 20˚C. What would have been the final temperature of the water?
serg [7]

Answer:

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Explanation:

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2 years ago
Read 2 more answers
A.) A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. If the titration required 1
nikklg [1K]

Answer:

A) 0.1225 M

B) 100.4 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)

Step 2: Calculate the reacting moles of NaOH

17.73 mL of 0.1036 M NaOH react. The reacting moles are:

0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol

Step 3: Calculate the reacting moles of HA

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.

Step 4: Calculate the molar concentration of HA

1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:

M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M

Step 5: Calculate the molar mass of HA

1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:

0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol

5 0
3 years ago
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