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PolarNik [594]
3 years ago
6

How many protons, and neutrons are in nuclide tin-115?

Chemistry
2 answers:
HACTEHA [7]3 years ago
8 0
50 protons
65 neutrons
maria [59]3 years ago
7 0

50 protons

65 nuetrons

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2.When copper (II) nitrate reacts with sodium hydroxide, copper (11) hydroxide is produced.
Ann [662]

Answer:

 

2Cu + S ~~~> Cu2S Copper (C) reacts with sulfur (S) to form copper sulfide as shown in the equation. A scientist adds 12.7 grams of Cu to 3.2 grams of S to start the reaction.

Explanation:

7 0
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Min : sec helium and nickel are classified as _____, which are pure substances that make up all kinds of matter
Sidana [21]

Helium which has a chemical formula of He is considered as a noble gas while Nickel which has a chemical formula of Ni is considered as metallic. Although both have different phases but they are both classified under the general category of “Elements”.

 

answer:

<span>elements</span>

4 0
3 years ago
What is the hyphen notation of nitrogen
vovikov84 [41]
<span> N-16

</span>7 protons & 9 neutrons
<span>7 + 9 = 16

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6 0
4 years ago
Imagine you pour some hot soup into a bowl. You place your hands on the outside of the bowl and feel that the bowl is very warm.
yuradex [85]

Um because of the energy in the heat? XD idk

5 0
3 years ago
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
3 years ago
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