A 50 ml sample of calcium hydroxide solution required 19.50 ml of 0.1050N HCL titration. Calculate the amount of calcium hydroxide in the 50 sample?

Answers

Answer 1

Answer: The reaction of calcium hydroxide (Ca(OH)2) and hydrochloric acid (HCl) is:
Ca(OH)2 + 2HCl --> CaCl2 + 2H2O. So the amount of calcium hydroxide in the sample is ((19.50ml HCl)/1000) x 0.1050 mol/L
From the balanced equation above: (1 mol Ca(OH)2 / 2 mol HCl)
Calculating the amount of Ca(OH)2 in the 50ml sample, multiply with the molar mass:
((19.50ml HCl)/1000) x 0.1050 mol/L x (1 mol Ca(OH)2 / 2 mol HCl) x 74.039 g/mol

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Given the balanced equation representing a reaction: 2na(s) + cl2(g) → 2nacl(s) + energy if 46 grams of na and 71 grams of cl2 react completely, what is the total mass of nacl produced? 1. 58.5 g 2. 117 g 3. 163 g 4. 234 g

Answers

Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.

How many grams of Cl are in 535 g of CaCl2?

Answers

m( $Ca Cl_(2)$ )=535 g
m(Cl)=?
The molar mass of $CaCl_(2)$ = 40.08 + 2 * 35.45 = 110.98
The atomic mass of Cl: 35.45 ( you can see it in the Periodic table)
110.98 g $CaCl_(2)$ - 35.45 * 2 g Cl
535 g $CaCl_(2)$ - x g Cl
Proportion:
110.98 : 535 = (35,45 * 2) : x
x = 535 * 35.45 * 2 / 110.98 = 37,931.5/110.98 = 341.79 g
Answer: m(Cl)=341.79 g

How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 percent by mass iron (II) nitrate solution react with excess aluminum metal? Show all of the work needed to solve this problem. 2Al (s) + 3Fe(NO3)2 (aq) yields 3Fe (s) + 2Al(NO3)3 (aq)

Answers

The balanced reaction is:

2Al (s) + 3Fe(NO3)2 (aq) = 3Fe (s) + 2Al(NO3)3 (aq)

We are given the amount of iron (II) nitrate solution with its purity. This will be the starting point of our calculation.

265 (.845) gram iron (II) nitrate (1 mol Fe(NO3)2 / 179.85 g Fe(NO3)2) ( 3 mol Fe / 3 mol Fe(NO3)2) (55.85 g Fe / 1 mol Fe) = 69.54 g Fe

The atomic mass of an element is calculated using the(1) atomic number and the ratios of its naturally occurring isotopes(2) atomic number and the half-lives of each of its isotopes(3) masses and the ratios of its naturally occurring isotopes(4) masses and the half-lives of each of its isotopes

Answers

The atomic mass of an element is calculated using its masses and the ratios of its naturally occurring isotopes. The answer is number 3. The rest of the choices do not answer the question above.

Answer:

yo slufEyospralryarlualrualruOruRhls

Elements in Group 17 form which ion?
A. 1
B. +1
C. -1
D. 0

Answers

Answer:

Elements in group 17 form C. -1 ions

It would b C. -1
hope this helps

Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg + N2 → Mg3N2 In a particular experiment, a 9.27-g sample of N2 reacts completely. The mass of Mg consumed is __________ g.

Answers

Mass is the amount of the substance present in the sample of the mixture, it is a quantitative measure. The mass of the magnesium consumed is 23.76 gm.

What is mass?

Mass of any substance is the product of the moles of the substance and the molar mass of the substance present in the sample mixture.

The reaction can be shown as,

$\rm 3 Mg + N_(2) \rightarrow Mg_(3)N_(2)$