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kiruha [24]
2 years ago
10

Elements with atomic numbers 112 and 114 have been produced and their iupac names are pending approval. however, an element that

would be put between these two elements on the periodic table has not yet been produced. if produced, this element will be identified by the symbol uut until an iupac name is approved.determine the charge of an uut nucleus. your response must include both the numerical value and the sign of the charg
Chemistry
1 answer:
victus00 [196]2 years ago
8 0
Answer is: +113.
Nihonium<span> is a </span>chemical element<span> with symbol </span>Nh<span> and </span>atomic number 113. It has 113 protons (<span>positive </span>electric charge<span> of +1</span>). P<span>rotons, beside neutrons (neutral charge), are in the </span>nucleus<span> of every </span><span>atom. Because only protons in nucleus have charge, charge of element-113 is +113.</span>
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jan is holding an ice cube. what causes the ice to melt? thermal energy from the ice is transferred to the air. thermal energy f
loris [4]

Answer: Ice is melting due to the transfer of thermal energy from Jan's hand to ice.

Explanation: The melting of ice is a physical change and is happening when the thermal energy from Jan's hand is transferred to ice. Due to this energy transfer, the particles of ice starts to move faster and hence, making the ice melt.

In this, the physical state of ice is changing from solid to liquid state.

H_2O(s)\rightleftharpoons H_2O(l)

8 0
2 years ago
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________ is the study of matter and the energy that causes matter to combine, break apart and recombine in everything living and
Kipish [7]

Answer: Emperical formula

Explanation:

8 0
2 years ago
A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of
allochka39001 [22]

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

8 0
2 years ago
A. A nurse practitioner prepares 500. mL of an IV of normal saline solution to be delivered at a rate of 80. mL/h. What is the i
nydimaria [60]
A. Quantity of saline = 500mL 
Rate of infusion = 80 mL / h 
Infusion time = Quantity / Rate = 500 mL / (80 mL/hr) = 6.25 hr 
b. Child weight = 72.6 lb = 32.93 kg 
Medrol to be given = 1.5 mg per kg 
Quantity of Medrol = 20 mg/mL 
Dosage available = 20 mg/mL / 1.5 mg/kg = 13.33 kg/mL 
Dosage according to body weight = 32.93 kg / 13.33 kg/mL = 2.47 mL
8 0
2 years ago
Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample
Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05\frac{g}{mol}.

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · \frac{1 mol Na^2SO^4}{142.05g Na^2SO^4} · \frac{4 mol O}{1 mol Na^2SO^4} ·\frac{6.022x10^2^3}{1 mol O}

After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.

3 0
2 years ago
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