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insens350 [35]
3 years ago
6

I need help ASAP k12 7th grade TGA 2.12

Chemistry
1 answer:
Rina8888 [55]3 years ago
7 0

Answer:

So you download the doc and do it.

Explanation:

So you download the doc and do it.

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How do I find the number of neutrons in an atom?
Vaselesa [24]

You need to subtract the number of protons from the mass number.

4 0
3 years ago
Read 2 more answers
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.

6 0
3 years ago
Iodine pentafluoride gas reacts with iodine fluoride gas producing iodine heptafluoride gas and iodine gas. What is the maximum
dybincka [34]

Answer:

63.45g of I₂ can be produced

Explanation:

IF₅ reacts with IF to produce IF₇ and I₂. The reaction is:

IF₅ + 2 IF → IF₇ + I₂

Moles of 10.0g of IF₅ (221.89g/mol):

10.0g IF₅ × (1mol / 221.89g) = <em>0.0451 moles of IF₅</em>

Using PV / RT = n, it is possible to find moles of 11.20L of IF, thus:

1atm×11.20L / 0.082atmL/molK × 273K = 0.500 moles of IF.

<em>At STP, pressure is 1atm, temperature is 273K and gas constant R is 0.082atmL/molK</em>

For a complete reaction of IF₅ you need:

0.0451 moles of IF₅ × (2 moles IF / 1 mole IF₅) = 0.902 moles of IF. As you have just 0.500 moles of IF, the IF is the limiting reactant.

2 moles of IF produce 1 mole of I₂. 0.500 moles of IF produce:

0.500mol IF ₓ ( 1 mol I₂ / 2 mol IF) = 0.250 mol I₂

As molar mass of I₂ is 253.81g/mol, mass of 0.250 mol I₂ are:

0.250mol I₂ ₓ (253.81g / mol) =

<h3>63.45g of I₂ can be produced</h3>
3 0
4 years ago
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
What is associated with an object that has gravitational potential energy
Molodets [167]

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s2.

8 0
3 years ago
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