Question

38 Calculate formula mass and the molar mass of \( \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \).

Ask by Moreno Bartlett. in Canada
Mar 18,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The formula mass and molar mass of \( \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \) is approximately \(147.01 \, \text{g/mol}\).

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Beyond the Answer

To calculate the formula mass and molar mass of \( \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \), we start by identifying the atomic masses of each element involved. Calcium (Ca) has a mass of approximately 40.08 g/mol, chlorine (Cl) about 35.45 g/mol, and hydrogen (H) and oxygen (O) are about 1.01 g/mol and 16.00 g/mol, respectively. For \( \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \): - Calcium: 1 x 40.08 = 40.08 g/mol - Chlorine: 2 x 35.45 = 70.90 g/mol - Water (from \( 2 \mathrm{H}_{2} \mathrm{O} \)): 2 x (2 x 1.01 + 16.00) = 2 x 18.02 = 36.04 g/mol Adding these together: 40.08 + 70.90 + 36.04 = 147.02 g/mol. So, the molar mass of \( \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \) is 147.02 g/mol. When it comes to real-life applications, calcium chloride is widely used not just in various chemical labs but also on roads in winter to manage ice, acting as an effective de-icing agent. Additionally, its hydrated form \( \mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \) is also utilized in food preservation and to keep moisture in packaged goods. Its versatility makes it a staple in many industries!

Related Questions

Complete each of the following nuclear decay equations by determining the mass number and atomic number of the following isotopes after they have emitted either a beta particle, an alpha particle or gamma radiation. Use the periodic table to assist you in identifying the remaining particle. a. \( \begin{array}{r}210 \\ 82 \\ \mathrm{~Pb}\end{array} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) 1. \( \begin{array}{lll}\frac{1}{\vdots} & \mathrm{~T} \\ \vdots & \rightarrow & 0 \gamma+ \\ \vdots & & 0\end{array} \) \( \qquad \) b. \( { }_{84}^{209} \mathrm{PO} \rightarrow{ }_{82}^{205} \mathrm{~Pb}+ \) \( \qquad \) j. \( { }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+ \) \( \qquad \) c. \( { }_{92}^{239} \mathrm{U} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) d. \( { }_{92}^{238} \mathrm{U} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) k. \( { }_{94}^{239} \mathrm{Pu} \rightarrow \quad{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) I. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) e. \( \begin{array}{r}228 \\ 93 \\ \mathrm{~Np}\end{array} \rightarrow \quad-1 \mathrm{e}+ \) \( \qquad \) f. \( \begin{array}{l}42 \\ 19\end{array} \mathrm{~K} \quad{ }_{19}^{42} \mathrm{~K}+ \) \( \qquad \) m. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) g. \( { }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) ก. \( { }_{4}^{9} \mathrm{Be} \rightarrow{ }_{4}^{9} \mathrm{Be}+ \) \( \qquad \)
12)(UCS-RS) Uma pessoa usou \( 34,2 \mathrm{~g} \) de sacarose \( \left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) \) para adoçar seu cafezinho. O volume de cafezinho adoçado na xícara foi de 50 mL . Qual foi a concentração da sacarose nesse cafezinho? ( \( \mathrm{C}=12 ; \mathrm{H}=1 ; \mathrm{O}=16 \) ) a) \( 0,5 \mathrm{~mol} / \mathrm{L} \) b) \( 1,0 \mathrm{~mol} / \mathrm{L} \) c) \( 1,5 \mathrm{~mol} / \mathrm{L} \) d) \( 2,0 \mathrm{~mol} / \mathrm{L} \) e) \( 2,5 \mathrm{~mol} / \mathrm{L} \) 13)(UFRR) Quantos gramas de sulfato de alumínio, \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \), são necessários para preparar 6 litros de uma solução 3 molar? (dados: \( \mathrm{Al}=27 ; \mathrm{S}=32 ; \mathrm{O}=16 \) ) a) 342 g . b) 615 g . c) 567 g . d) 765 g . e) 6156 g . 14)(VUNESP) - Com o objetivo de diminuir a incidência de cáries na população, em muitas cidades adiciona-se fluoreto de sódio ( NaF ) à água distribuída pelas estações de tratamento, de modo a obter uma concentração de \( 2,0 \times 10^{-5} \) \( \mathrm{mol} / \mathrm{L} \). Com base neste valor e dadas as massas molares em g/mol: \( \mathrm{Na}=23 \) e \( \mathrm{F}=19 \), podemos dizer que a massa do sal contida em 500 mL desta solução é: a) \( 4,2 \times 10^{-1} \mathrm{~g} \). b) \( 8,4 \times 10^{-1} \mathrm{~g} \). c) \( 4,2 \times 10^{-4} \mathrm{~g} \). d) \( 6,1 \times 10^{-4} \mathrm{~g} \). e) \( 8,4 \times 10^{-4} \mathrm{~g} \). 15)(Mackenzie-SP) Qual é, respectivamente, a molaridade do íon \( \mathrm{Mg}^{2+} \) e a do \( \left(\mathrm{PO}_{4}\right)^{3-} \) numa solução 0,4 molar de \( \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2} \) ? a) 2 e 3 b) 3 e 2 c) 2,4 e 2,4 d) 0,4 e 0,4 e) 1,2 e 0,8
Chemistry Brazil Mar 18, 2025

Latest Chemistry Questions

Complete each of the following nuclear decay equations by determining the mass number and atomic number of the following isotopes after they have emitted either a beta particle, an alpha particle or gamma radiation. Use the periodic table to assist you in identifying the remaining particle. a. \( \begin{array}{r}210 \\ 82 \\ \mathrm{~Pb}\end{array} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) 1. \( \begin{array}{lll}\frac{1}{\vdots} & \mathrm{~T} \\ \vdots & \rightarrow & 0 \gamma+ \\ \vdots & & 0\end{array} \) \( \qquad \) b. \( { }_{84}^{209} \mathrm{PO} \rightarrow{ }_{82}^{205} \mathrm{~Pb}+ \) \( \qquad \) j. \( { }_{90}^{234} \mathrm{Th} \rightarrow{ }_{91}^{234} \mathrm{~Pa}+ \) \( \qquad \) c. \( { }_{92}^{239} \mathrm{U} \rightarrow \quad{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) d. \( { }_{92}^{238} \mathrm{U} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) k. \( { }_{94}^{239} \mathrm{Pu} \rightarrow \quad{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) I. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) e. \( \begin{array}{r}228 \\ 93 \\ \mathrm{~Np}\end{array} \rightarrow \quad-1 \mathrm{e}+ \) \( \qquad \) f. \( \begin{array}{l}42 \\ 19\end{array} \mathrm{~K} \quad{ }_{19}^{42} \mathrm{~K}+ \) \( \qquad \) m. \( { }_{13}^{27} \mathrm{Al} \rightarrow{ }_{-1}^{0} \mathrm{e}+ \) \( \qquad \) g. \( { }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{2}^{4} \mathrm{He}+ \) \( \qquad \) ก. \( { }_{4}^{9} \mathrm{Be} \rightarrow{ }_{4}^{9} \mathrm{Be}+ \) \( \qquad \)
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy