What is a spelling bee​

Answers

Answer 1
Answer:

Answer: a contest in which you are eliminated if you fail to spell a word correctly.

Answer 2
Answer: a contest where you attempt to spell a series of words that you are given by the judges

Related Questions

Electrons, protons, and neutrons are three subatomic particles of an atom. If the atom undergoes a change and becomes a positive ion, which subatomic particle is lost in this process? Neutron Proton Electron
ASAP pleaseee and Thankyou​​
How does the Atomic theoryimpact our understanding of the universe?
Which layer of the Basement Membrane of epithelial tissues is produced and secreted by cells of the underlying connective tissue? (a) Reticular Lamina (b) Lamellar Lamina (c) Papillary Lamina (d) Basal Lamina
What would be the best way to measure Carbon Dioxide in a Soft Drink can?And yes, we were going to do it by taste.

The pH scale for acids and bases ranges from

Answers

0-14
0-6 is an acid
8-14 is a base
7 is neutral

Put the elements potassium iron bromine and krypton in order of increasing electro negativity

Answers

Answer:

K < Fe < Br < Kr

Explanation:

Given chemical elements;

      Potassium

      Iron

      Bromine

      Krypton

Problem; arrange in order of increasing electronegativity

Solution:

The electronegativity of an element measures the relative tendency with which the atoms of the element attracts valence electrons in the chemical bond.

 To solve this problem, we are going to use values of electronegativity from the Pauling's electronegativity table of elements;

       For;

            Potassium  = 0.8

            Iron = 1.83

            Bromine  = 2.8

            Krypton = 3.0

So;

  K < Fe < Br < Kr

The standard enthalpy change for the reaction of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH∘ = -227.8 kJ . Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO4(aq) (in kilojoules per mole). [For H2O(l),ΔH∘f = -285.8kJ/mol].

Answers

Answer:

-909.3KJ/mole

Explanation:

The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:

ΔH = Σ ΔHf products - Σ ΔHf reactants

Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.

From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.

Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:

We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]

We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.

We then proceed to the second stage.

Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.

We go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].

We already know that the ΔH here equals -98.9KJ.

Hence, -98.9 = y + 296.8

y = -296.8KJ - 98.9KJ = -395.7KJ

We now proceed to the final part of the calculation which ironically comes first in the series of sentences.

Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.

Mathematically, we go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].

Now, we know that the ΔH for this particular reaction is -227.8KJ

We then proceed to to open the bracket.

-227.8 = z - (-395.7 - 285.8)

-227.8 = z - ( -681.5)

-227.8 = z + 681.5

z = -227.8-681.5 = -909.3KJ

Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol

What volume, in Liters, of a 2.00 M KCl solution contains 2.5kg of KCl? Molar mass of KCl= 74.55g/mole

Answers

According to the molar concentration, volume in liters of 2 M KCl solution is 16.76 L.

What is molar concentration?

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.

In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.

Volume required is obtained by molarity formula as, 2×74.55/2500=1/V=0.05964

∴V=1/0.05964=16.67 L.

Thus, 16.67 L of  2 M KCl solution contains 2.5 kg of KCl.

Learn more about molar concentration,here:

brainly.com/question/21841645

#SPJ2

C_(m)=(n)/(V_(r))\n\nn=(m_(s))/(M)\n\nC_(m)=(m_(s))/(MV_(r)) \ \ \ |*MV_(r)\n\nC_(m)MV_(r)=m_(s) \ \ \ |:C_(m)M\n\nV_(r)=(m_(s))/(C_(m)M)\n\n\nm=2.5kg=2500g\nM=74.55(g)/(mol)\nC_(m)=2(mol)/(L)\n\nV_(r)=(2500g)/(2(mol)/(L)*74.55(g)/(mol))\approx16.77L

The normal boiling point corresponds to the temperature at which both the liquid and gas are under the standard atmospheric pressure of 1 atm.TRUE

FALSE

Answers

TRUE. (Lorenzo Romano Amadeo Carlo Avogadro) Ideal Gas Law that defined as one in which all collisions between atoms or molecules are perfectly elastic in which there are no intermolecular attractive forces.  In such a gas, all the internal energy is the form of kinetic energy and any change in internal energy is accompanied by a change in temperature. That characterized by three state variables: absolute pressure (P) = 1 atm, volume (V) = 22.4 L and absolute temperature (T) = 273 K.

In Rutherford’s gold foil experiment, many of the alpha particles passed straight through the gold foil, completely undeflected. What did Rutherford conclude from this observation? A. Alpha particles have a positive charge.

B. Electrons have a negative charge.

C. An atom is made up of mostly empty space.

D. There is a dense area within an atom that has a negative charge.

Answers

Answer: C

Explanation:

Rutherford's Gold Foil Experiment

These were his conclusions.

  • Aimed beam of alpha particles at a thin piece of gold foil and found most of the particles passed through the gold foil without being scattered, except for a few particles.
  • He found the volume of an atom is mostly empty space.
  • He proposed there is a heavy, positively charged body at the center of each atom that contains most of the mass.
  • Electrons are throughout the atom.
  • Number of electrons outside the nucleus equals the number of protons in the nucleus. Atom is electrically neutral.