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Papessa [141]
3 years ago
14

The noble gases are inert. This means they a. exist as gases at room temperature. b. undergo many chemical reactions. c. lose an

d gain electrons easily. d. undergo very few chemical reactions.
Chemistry
1 answer:
eduard3 years ago
8 0

Answer:

D

Explanation:

The noble gases are said to be inert because they undergo very few chemical reactions. They enjoy their status by having completely filled electron shells and do not have any reason whatsoever to go into chemical reactions. Most elements otherwise go into chemical reactions principally to enjoy the stability of these elements.

Noble gases can be found in the last group of the periodic table.

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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
Please help me to answer this question.<br>i have to submit tomorrow.​
ratelena [41]

Answer:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

4 0
3 years ago
How does evaporation show interactions among the hydrosphere and atmosphere?
Vladimir79 [104]

Answer:

water evaporates from the ocean into atmosphere. water vapour condenses to form clouds. clouds produce rain. rainwater needed for plant growth.

Explanation:

6 0
3 years ago
Read 2 more answers
At a certain temperature, the solubility of strontium arsenate, sr3(aso4)2, is 0.0560 g/l. what is the ksp of this salt at this
REY [17]
The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles 
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq) 
Ksp = (Sr++)^3(As04^-3)^2 
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4 
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5 0
4 years ago
Draw the ground-state electronic configuration for Mg.
vichka [17]

Answer:

(Ne)3s2

Explanation:

6 0
3 years ago
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