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2.13 g (6.02 x 1023 atoms) / (1.01 + 79.91) = 1.58 x 1022 atoms

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How many moles are in 186 grams of HBr?

For this you need the atomic (molecular) mass of HBr. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. HBr= 81.0 grams186 grams HBr / (81.0 grams) =2.30 moles HBr


How many grams in 4.23 m mol of HBr?

The answer is 0,3422 grams.


How many grams of aluminum bromide are formed by the reaction of 121 g of HBr according to the equation?

The balanced chemical equation is: 6HBr + Al2(SO4)3 -> 2AlBr3 + 3SO3 + 6H2O. To find the grams of aluminum bromide formed, you need to calculate the molar ratio of HBr to AlBr3 and then convert 121g of HBr to grams of AlBr3.


How many atoms are in HBr?

There is one hydrogen atom and one bromine atom in one molecule of HBr.


Is HBr a molecular compound?

Yes, HBr is a molecular compound. It consists of covalent bonds between hydrogen and bromine atoms, forming molecules of HBr.


Which is least polar HF HBr HCl or Hbr?

HF is the least polar among these molecules because it has the smallest difference in electronegativity between the hydrogen and the fluorine atoms.


How many grams HBr are required to produce 150g AlBr3 in the equation 2Al plus 6HBr producing 2AlBr3 and 3H2?

Given the balanced equation2Al + 6HBr --> 2AlBr3 + 3H2In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr----------- 266.6g AlBr3 * 2 molecules AlBr3


What kind of bond is HBr?

HBr is a polar covalent bond because hydrogen and bromine have different electronegativities, leading to an unequal sharing of electrons between the two atoms.


Is HBr linear or trigonal pyramidal?

As it is a diatomic molecule it can only be linear.


Is HBr a dipole dipole?

Yes, HBr is a polar molecule with a significant difference in electronegativity between hydrogen and bromine atoms. This results in a permanent dipole moment, making it exhibit dipole-dipole interactions with other polar molecules.


How many grams of solid sodium hydroxide would need to be added to completely neutralize 35.0 mL of 1.45 M HBr?

This is a titration question: we want to have the same number of hydroxide ions as hydroxide ions so that they will form water and the pH will be neutral. In chemistry, we count atoms and molecules in moles, and we can calculate how many moles of HBr we have, because concentration in molarity is the number of moles divided by the volume in liters... M = moles/V. We plug in what we got: 1.45M = moles/0.0350L, and solve for moles: 0.0508 moles. Now we know we need 0.0508 moles of NaOH, whose molecular weight is 40g/mole. MW x moles = grams, so (40g/mole)(0.0508 moles) = 2.03 g of NaOH.


What is the net ionic equation for HBr and H2O?

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