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Brrunno [24]
3 years ago
9

What is the atomic number of an element that has 43 protons and 50 neutrons?

Chemistry
1 answer:
zimovet [89]3 years ago
8 0
The atomic number is based on the number of protons, so the atomic number would be 43
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Identify each element below, and give the symbols of the other elements in its group:
lozanna [386]

Answer:

Answer in explanation

Explanation:

Argon has 18 electrons. So to get the element in question, we only need to add 18 to the number of the filled electrons.

a. Germanium, atomic number 32

Other group members:

Silicon Si , Carbon C , Tin Sn , Lead Pb and Flerovium Fl

b. Cobalt , atomic number 27

Other group members:

Rhodium Rh , Iridium Ir and Meitnerium Mt

c. Technetium , atomic number 43

Krypton is element 36

Other group members are :

Manganese Mn , Rhenium Re and Bohrium Bh

7 0
4 years ago
Use the Gizmo to estimate the age of each of the objects below. For these questions, each second in the Gizmo represents 1,000 y
ziro4ka [17]

Answer:

Check the explanation

Explanation:

AT = A0 e(-T/H)

... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.

Substituting what we know, we get...

0.71 = (1) e(-T/5730)

Solve for T...

loge(0.71) = -T/5730

T = -loge(0.71)(5730)

T = 1962 (conservatively rounded, T = 2000)

similarly for all

for aboriginal charcoal

0.28 = (1) e(-T/5730)

Solve for T...

loge(0.28) = -T/5730

T = -loge(0.28)(5730)

T = 7294 (conservatively rounded, T = 7000)

for mayan headdress

0.89 = (1) e(-T/5730)

Solve for T...

loge(0.89) = -T/5730

T = -loge(0.89)(5730)

T = 667 (conservatively rounded, T = 700)

for neanderthal

0.05 = (1) e(-T/5730)

Solve for T...

loge(0.05) = -T/5730

T = -loge(0.05)(5730)

T = 17165 (conservatively rounded, T = 17000)

7 0
3 years ago
Which metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe? auau cucu mnmn agag
Cerrena [4.2K]

Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.

Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.

Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Learn more about electrode here:

brainly.com/question/17060277

#SPJ4

3 0
2 years ago
A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed
MariettaO [177]

Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: <em>E°cell=Ereduction - Eoxidation</em>

Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

<em>E°cell= 1.56 V</em>

Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

6 0
3 years ago
Using the following equation 5KNO2+2KMnO4+3H2SO4=5KNO3+2MnSO4+K2SO4+3H20. Starting with 2.47 grams of KNO2 and excess KMnO4 how
Aleonysh [2.5K]

Given equation : 5KNO_{2} + 2KMnO_{4} + 3H_{2}SO_{4}\rightarrow 5KNO_{3} + 2MnSO_{4} + K_{2}SO_{4} + 3H_{2}O

Given information = 2.47 grams KNO2 and excess KMnO4 and we need to find grams of water (H2O).

Since KMnO4 is in excess, so grams of water(H2O) can be calculated using grams of KNO2 with the help of stoichiometry.

To find grams of water(H2O) from grams of KNO2 , we need to follow three steps.

Step 1. Convert 2.47 grams of KNO2 to moles of KNO2.

Moles = \frac{grams}{molar mass}

Molar mass of KNO2 = 85.10 g/mol

Moles = 2.47 gram KNO2\times \frac{1 mol KNO2}{85.10 gram KNO2}

Moles = 0.0290 mol KNO2

Step 2. Convert moles of KNO2 to moles of H2O using mole ratio.

Mole ratio are the coefficient present in front of the compound in the balanced equation.

Mole ratio of KNO2 : H2O is 5 : 3 (5 coefficient of KNO2 and 3 coefficient of H2O)

0.0290 mol KNO2\times \frac{3 mol H2O}{5 mol KNO2}

Mole = 0.0174 mol H2O

Step 3. Convert mole of H2O to grams of H2O

Grams = Moles X molar mass

Molar mass of H2O = 18.00 g/mol

Grams = 0.0174 mol H2O\times \frac{18 g H2O}{1 mol H2O}

Grams of water = 0.313 grams H2O

Summary : The above three steps can also be done in a singe setup as shown below.

2.47 gram KNO2\times \frac{(1 mol KNO2)}{(85.10 gram KNO2)}\times \frac{(3 mol H2O)}{(5 mol KNO2)}\times \frac{(18.00 gram H2O)}{(1mol H2O)}

In the above setup similar units get cancelled out and we will get grams of H2O as 0.313 grams water (H2O)

8 0
3 years ago
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