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guajiro [1.7K]
4 years ago
6

What causes airbags to deploy in an accident

Chemistry
2 answers:
HACTEHA [7]4 years ago
6 0
The airbag sensor being struck
Law Incorporation [45]4 years ago
5 0
If the care is impacted by something over 25 miles per hour
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Calculate the molar mass of nitrogen gas if 0.250 g of the gas occupies 46.65 ml at stp
Mashutka [201]
<span>pre-1982 definition STP: 120 g/mol post-1982 definition STP: 122 g/mol The answer to this question depends upon which definition of STP you're using. The definition changed in 1982 from 273.15 K at 1 atmosphere to 273.15 K at 10000 pascals. As a result the molar volume of a gas at STP changed from 22.4 L/mol to 22.7 L/mol. So let's calculate the answer using both definitions and see if your text book is 35 years obsolete. First, determine the number of moles of gas you have. Do this by dividing the volume you have by the molar volume. So pre-1982: 0.04665 / 22.4 = 0.002082589 mol post-1982: 0.04665 / 22.7 = 0.002055066 mol Now divide the mass you have by the number of moles. pre-1982: 0.250 g / 0.002082589 mol = 120.0428725 g/mol post-1982: 0.250 g / 0.002055066 mol = 121.6505895 g/mol Finally, round to 3 significant figures: pre-1982: 120 g/mol post-1982: 122 g/mol These figures are insanely large for nitrogen gas. So let's see if our input data is reasonable. Looking up the density of nitrogen gas at STP, I get a value of 1.251 grams per liter. The value of 0.250 grams in the problem would then imply a volume of about one fifth of a liter, or about 200 mL. That is over 4 times the volume given of 46.65 mL. So the verbiage in the question mentioning "nitrogen gas" is inaccurate at best. I see several possibilities. 1. The word "nitrogen" was pulled out of thin air and should be replaced with "an unknown" 2. The measurements given are incorrect and should be corrected. In any case, if #1 above is the correct reason, then you need to pick the answer based upon which definition of STP your textbook is using.</span>
6 0
3 years ago
Express the sum of 9.78 g and 7.0 g using the correct number of significant digits.
kvv77 [185]

Answer:

The correct answer is 16.8 g (option C)

Explanation:

Step 1: Data given

9.78 grams + 7.0 grams

Rule significant numbers say:

⇒ Non-zero digits are always significant.

⇒ Any zeros between two significant digits are significant.

⇒ A final zero or trailing zeros in the decimal portion ONLY are significant.

For addition and subtraction, look at the places to the decimal point.  Add or subtract in the normal way, then round the answer to the LEAST number of places to the decimal point of any number in the problem.

9.78 g+ 7.0 g = 16.78 grams

Significant numbers = 1 decimal

16.78 ⇒ 16.8

The correct answer is 16.8 g (option C)

4 0
3 years ago
Read 2 more answers
What happens in a single-replacement reaction?
antoniya [11.8K]

Answer:

D

Explanation:

A single-replacement reaction occurs when one element replace another in a single compound.

8 0
3 years ago
If you put an egg on a sidewalk on a hot day in July and try to cook it you are using?
Zepler [3.9K]

Answer:

  • <em><u>Passive solar energy</u></em>

Explanation:

First of all, you must know that you if you put an egg on a sidewalk you are dealing with energy from the Sun, i.e. solar energy, while geothermal energy is energy that comes from the inner of the Earth and biomass energy comes from plant or animal material.

The term passive solar energy refers to the fact that the energy of the sun is used directly for the intended task, which in this case is to cook the egg.

The term active solar energy refers to the fact that the energy of the Sun is converted into a  different form of energy and then used for your purpose. For instance, if the energy of the Sun were used to produce electricity and then this electricity used to cook the egg, you would be using an acitve solar energy.

5 0
3 years ago
The basic particle from which all elements are made; the smallest particle of an element that has the properties of that element
Anastasy [175]

Answer:

<h2><u>Atom.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u></h2>

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3 years ago
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