0
What else can I help you with?
How many grams of Br are in 595 g of CaBr2?
To find the grams of bromine (Br) in 595 g of calcium bromide (CaBr2), first determine the molar mass of CaBr2. The molar mass is approximately 40.08 g/mol for Ca and 79.90 g/mol for Br, giving a total of about 199.89 g/mol for CaBr2. Since there are two bromine atoms in each formula unit, the mass of bromine in CaBr2 is about 159.80 g (2 × 79.90 g). Thus, the mass of Br in 595 g of CaBr2 can be calculated as follows: (159.80 g Br / 199.89 g CaBr2) × 595 g CaBr2 ≈ 477.23 g of Br.
What is reducing agent in Ca Br2--CaBr2?
In the reaction Ca + Br2 → CaBr2, calcium acts as the reducing agent because it undergoes oxidation by losing electrons to bromine. This results in the formation of calcium bromide (CaBr2).
What is the balanced equation of the reactants MnSO4 and Na2CO3?
The balanced equation for the reaction between MnSO4 and Na2CO3 is: MnSO4 + Na2CO3 -> MnCO3 + Na2SO4.
Chemical formula of washing soda?
the chemical formula of washing soda is Na2CO3 (Sodium carbonate).
How value of x can be calculated in the formula Na2CO3?
With great difficulty - since there is no x in Na2CO3.
How many moles of CaBr are in 5.0 grams of CaBr2?
To find the number of moles of CaBr2 in 5.0 grams, you first need to calculate the molar mass of CaBr2. The molar mass of CaBr2 is 200.8 g/mol. Divide the given mass by the molar mass to find the number of moles: 5.0 g / 200.8 g/mol = 0.025 moles of CaBr2. Since there is one mole of CaBr2 for every two moles of CaBr, you have half of that amount in moles of CaBr: 0.025 moles / 2 = 0.0125 moles of CaBr.
What is the proper formula for calcium bromide?
The chemical formula for calcium bromide is CaBr2.
How many moles of CaBr2 are in 5.0 grams of CaBr2?
Well, darling, if you want to know how many moles of CaBr2 are in 5.0 grams, you just need to divide the mass by the molar mass of CaBr2. The molar mass of CaBr2 is approximately 199.89 g/mol, so 5.0 grams of CaBr2 is roughly 0.025 moles. Hope that helps, sugar!
How many grams of CaBr2 are there in 0.31moles of CaBr2?
To find the number of grams of CaBr2 in 0.31 moles, you first calculate the molar mass of CaBr2, which is approximately 199.89 g/mol. Then, you can multiply the number of moles (0.31 moles) by the molar mass to find the grams: 0.31 moles x 199.89 g/mol = 61.97 grams of CaBr2.
Can you balance this equation of Ca plus Br2 equals CaBr2?
Ca + Br2 = CaBr2 doesn't need to be balanced.
Is a solution of CaBr2 acidic basic or neutral?
A solution of CaBr2 is neutral. When CaBr2 dissolves in water, it dissociates into calcium ions (Ca2+) and bromide ions (Br-), which do not significantly affect the pH of the solution.
How many grams of Br are in 595 g of CaBr2?
To find the grams of bromine (Br) in 595 g of calcium bromide (CaBr2), first determine the molar mass of CaBr2. The molar mass is approximately 40.08 g/mol for Ca and 79.90 g/mol for Br, giving a total of about 199.89 g/mol for CaBr2. Since there are two bromine atoms in each formula unit, the mass of bromine in CaBr2 is about 159.80 g (2 × 79.90 g). Thus, the mass of Br in 595 g of CaBr2 can be calculated as follows: (159.80 g Br / 199.89 g CaBr2) × 595 g CaBr2 ≈ 477.23 g of Br.
What is the name of ionic compound CaBr2?
The answer to this question is Calcium (Ca) Br2 (-ide) Bromide. Put them together, you get Calcium Bromide.
How are 0.50m Na2CO3 and 0.50M Na2CO3 different?
difference between 0.50mol na2co3 anf 0.50 M of na2co3
Convert 14 moles of caBr2 to grams?
1st you must find the molar mass of CaBr2. Ca 40.08g * (number of moles in the compound) 1 Br 79.90g * 2 This give you the amount (g) in 1 mole of CaBr2. Multiply by 14 and you get the answer.
What solute particles are present in an aqueous solution of CaBr2?
In an aqueous solution of CaBr2, the solute particles are Ca2+ cations and Br- anions. When dissolved in water, CaBr2 dissociates into these ions, which are responsible for conducting electricity and other properties of the solution.
What is reducing agent in Ca Br2--CaBr2?
In the reaction Ca + Br2 → CaBr2, calcium acts as the reducing agent because it undergoes oxidation by losing electrons to bromine. This results in the formation of calcium bromide (CaBr2).
