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What would you multiply atoms of boron by to get the units grams of boron?
There is no direct relationship between atoms of boronand grams of boron. Use Avogadro's number to convert atoms to moles, and the atomic mass to convert moles to grams.Since you are converting from atoms B, this goes in the denominator (on the bottom) of the first factor. You want to end up in units of grams of B, so this goes in the numerator (on the top) of the last factor.atoms B1.00 mole B10.8 gram = g B6.02E+23 atom B1.00 mole BNote that the units atoms boron "cancel out" in the first factor and you are left in units of moles. Moles cancel outin the second factor and the final units are grams boron.
A plus 2B yields 3C If he starts with 5 grams of A and 8 grams of B is B the limiting reactant and why?
Yes, B is the limiting reactant because for every one mole of A reacting with two moles of B, the ideal ratio is 1:2. Since you have more moles of B available (8g is more than 5g), the reaction will consume all 5g of A before it can use up all 8g of B.
What is mole-mass stoichiometry?
The number of moles in the equation of B DIVIDED BY the number of moles in the equation of A ANSWER TIMES the molar mass of B OVER 1. A is the the known compound, B is the one your trying to find out.
How many moles are in 97.2 grams of boron?
To do this, you need to know the atomic weight of the element you're dealing with, found on any periodic table. The atomic weight is the mass in grams of the element in one mole - this will provide you with a conversion factor. So take the measurement in grams and divide it by the atomic weight to convert to moles. Really what you're doing is multiplying the number by 1 mole, and dividing it by the equivalent of one mole, the atomic weight. That's the thought process behind unit analysis and how you get your "units to cancel".In this case, the answer is about 8.99 moles B.
What is the first step to most Stoichiometry problems?
The first step in most stoichiometry problems is to write a balanced chemical equation for the reaction you are investigating. This balanced equation is essential for determining the mole ratios between reactants and products, which are critical for solving stoichiometry problems.
How many B atoms are there in 2.47 moles of B?
The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / moleTo convert moles to atoms:atoms B = 2.47 mol B6.02 x 1023 atoms B = 1.49E+24 atoms B1 mol BMultiply by atoms per mole. Moles cancel out.
What steps should be used to determine the empirical formula of a compound from its mass percent composition?
A) Convert each mass to grams. B) Divide each molar amount by the ...Top answer: B) Divide each molar amount by the smallest molar amount
What is mole mass stoichiometry?
The number of moles in the equation of B DIVIDED BY the number of moles in the equation of A ANSWER TIMES the molar mass of B OVER 1. A is the the known compound, B is the one your trying to find out.
How many grams of S are there in a sample of S that contains the same number of moles as a 96.3 gram sample of B?
The atomic weight of B is 10.8 g/mol1. Convert grams of B to moles of B:moles B = 96.3 g B1 mol = 8.91 mol B10.8 gMultiply by moles per gram. Grams cancel out.The atomic weight of S is 32.1 g/mol2. To convert 8.91 moles of S to grams of S:grams S = 8.91 mol S32.1 g = 286 g S1 molMultiply by grams per mole. Moles cancel out.Note that since the atomic weight of S is larger than the atomic weight of B, the mass of the same number of moles is also larger.
How many moles of AL(NO3)3 will be produced when 0.75 moleof AGNO3 reacts?
Al + 3AgNO3 ==> Al(NO3)3 + 3Ag0.75 moles AgNO3 will produce 1/3 x 0.75 moles of Al(NO3)3 b/c of the mole ratio in the balanced eq.1/3 x 0.75 = 0.25 moles Al(NO3)3 will be produced.
How many moles of fluorine in 3 moles of borontrifluoride?
Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.
What are the risk factors of a mole?
GENERALLY, MOLES DO NOT present a problem, they are found on various parts of the body. HOWEVER, moles can become a form of skin cancer and need to be checked on occasion. The rule of thumb suggested by dermatologists when checking a mole is known as A B C. A: Asymmetry- The mole is not symmetical. B: Borders- The mole has uneven or jagged edges. C: Color- Any change in color. If a mole meets the ABC criteria, then it should be checked by a doctor, as this may suggest melanoma. If there is no change in a mole, then it's just a mole.
If one knows the mass and molar mass of reactant A and the molar mass of product D in a chemical reaction one can determine the mass of product D produced by using the?
Balanced chemical equation along with the stoichiometric ratios derived from that chemical reaction. A + B --> 2C mass of A * 1/molar mass of A = moles of A Moles of A * 2 moles of C/mole of A = moles of C Moles of C * molar mass of C = mass of C Also, you must think about limiting reagents, because if there is not enought reactant B to react with the amount of reactant A then the amount of reactant B will limit the production of product C!
Convert ratio to percent?
To convert ratio to percentage use the following method for a ratio where a:b means the same as a/b, then percentage = ( a / ( 1 - a ) ) x 100 so for a/b = 0.37 then percentage = (0.37 / (1 + 0.37)) x 100 = 27% a/b = 0.5, % = 0.33 a/b = 1, % = 0.5
What would you multiply atoms of boron by to get the units grams of boron?
There is no direct relationship between atoms of boronand grams of boron. Use Avogadro's number to convert atoms to moles, and the atomic mass to convert moles to grams.Since you are converting from atoms B, this goes in the denominator (on the bottom) of the first factor. You want to end up in units of grams of B, so this goes in the numerator (on the top) of the last factor.atoms B1.00 mole B10.8 gram = g B6.02E+23 atom B1.00 mole BNote that the units atoms boron "cancel out" in the first factor and you are left in units of moles. Moles cancel outin the second factor and the final units are grams boron.
A plus 2B yields 3C If he starts with 5 grams of A and 8 grams of B is B the limiting reactant and why?
Yes, B is the limiting reactant because for every one mole of A reacting with two moles of B, the ideal ratio is 1:2. Since you have more moles of B available (8g is more than 5g), the reaction will consume all 5g of A before it can use up all 8g of B.
What mass of sodium oxide could be made from 115grams of sodium?
Balanced equation: 2Na(s) + O2(g) ---> Na2O(s)moles Na = 115 g x 1 mol/23 g = 5 molesmoles Na2O possible = 1/2 x 5 moles = 2.5 moles (b/c the mole ratio of Na2O:Na is 1:2)mass Na2O possible = 2.5 moles x 62 g/mole = 155 grams (assuming O2 is not limiting)
