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larisa86 [58]
3 years ago
9

Elements in a given period of the periodic table contain the same number of

Chemistry
1 answer:
leva [86]3 years ago
4 0
Its C, Outermost Electrons.
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What is the acceleration of a 50kg object pushed with a force of 500 Newton’s
olasank [31]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m} \\

f is the force

m is the mass

From the question we have

a =  \frac{500}{50}  =  \frac{50}{5}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

5 0
3 years ago
Nitrogen increases volume during what physical change
atroni [7]
Effects of changes in volume in a reversible reaction in a chemical equilibrium can be predicted using Le Chatelier's Principle. I think this might be the answer, I hope it helps.
4 0
3 years ago
What is nuclear energy ​
blsea [12.9K]

Answer:

Nuclear energy comes from splitting atoms in a reactor to heat water into steam, turn a turbine and generate electricity.

Explanation:

Ninety-five nuclear reactors in 29 states generate nearly 20 percent of the nation's electricity, all without carbon emissions because reactors use uranium, not fossil fuels.

// have a great day //

5 0
3 years ago
Read 2 more answers
How many moles are 5.55 x 102 atoms of Mg?<br> M2) an&gt; NU X 02.0
cupoosta [38]

Explanation:

number of atoms = moles × avegadro number

so 5.55 × 10^2 = moles × 6.023 × 10^23

moles = 5.55 × 10^2 ÷ 6.023 × 10^23 = 9.214×10−22 moles

4 0
3 years ago
PLEASE HELP!! I REALLY NEED HELP
Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

<h2><u><em>56-57: NaCl</em></u></h2>

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na):   \frac{22.99}{58.443} = .393

2(Cl): \frac{35.453}{58.443}= .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

<h2><u>58-60 </u>K_{2} CO_{3}<u /></h2>

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: \frac{78.196}{138.204}= .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: \frac{12.011}{138.204}= .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: \frac{47.997}{138.204}= .347 <u>Step 3</u>: (.347)(100) = 34.7%

<h2>61-62 Fe_{3} O_{4}</h2>

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: \frac{167.535}{231.531}= .724 Step 3: (.724)(100)= 72.4%

2: O : \frac{63.996}{231.531}= .276 Step 3: (.276)(100) = 27.6%

<h2>63-65 C_{3}H_{5}(OH)_{3}</h2>

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: \frac{36.033}{92.094}= .391 Step 3: (.391)(100) = 39.1%

2: H: \frac{8.064}{92.094}= .088 step 3: (.088)(100) = 8.8%

2: O: \frac{47.997}{92.094} = .521 step 3: (.521)(100) = 52.1%

3 0
3 years ago
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