Question
upstudy study bank question image url

The second step in converting between the mass of a compound and moles of a compound is to

Ask by Ellis Reese. in the United States
Mar 18,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The second step is to use the molar mass of the compound.

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

Bonus Knowledge

determine the molar mass of the compound. The molar mass tells you how much one mole of that substance weighs in grams and is calculated by adding up the atomic masses of all the atoms present in the compound. For example, in water (H₂O), you would sum the atomic masses of hydrogen (1 g/mol) and oxygen (16 g/mol) to find that the molar mass is 18 g/mol. Next, you use the molar mass to facilitate conversions between mass and moles. If you know the mass of your compound, divide that mass by the molar mass to find the number of moles. Conversely, if you want to find the mass from a specific number of moles, you simply multiply the number of moles by the molar mass. It's like your own little math adventure!

Related Questions

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