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Delvig [45]
3 years ago
14

What is the acceleration of an object if it goes from a velocity of +25m/s to rest in 5.0s?

Chemistry
1 answer:
kari74 [83]3 years ago
5 0

Vf = Vi + at

0 m/s = 25 m/s + a(5sec)

-5 m/s2 = acceleration

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Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as
FrozenT [24]

Answer:

0.100 M AlCl₃

Explanation:

The variation of boiling point by the addition of a nonvolatile solute is called ebullioscopy, and the temperature variation is calculated by:

ΔT = W.i

Where W = nsolute/msolvent, and i is the Van't Hoff factor. Because all the substances have the same molarity, n is equal for all of them.

i = final particles/initial particles

C₆H₁₂O₆ don't dissociate, so final particles = initial particles => i = 1;

AlCl₃ dissociates at Al⁺³ and 3Cl⁻, so has 4 final particles and 1 initial particle, i = 4/1 = 4;

NaCl dissociates at Na⁺ and Cl⁻ so has 2 final particles and 1 initial particle, i = 2/1 = 2;

MgCl₂ dissociates at Mg⁺² and 2Cl⁻, so has 3 final particles and 1 initial particle, i = 3/1 = 3.

So, the solution with AlCl₃ will have the highest ΔT, and because of that the highest boiling point.

8 0
3 years ago
Help me answer this chemistry question. thank you!
gayaneshka [121]

Answer:

krypton,radon, and silicon

6 0
3 years ago
What does the atomic number of an element tell you?
stiv31 [10]
The atomic number of an element tells you the number of protons that an atom of the given element has.
7 0
3 years ago
What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3?
Anvisha [2.4K]

Answer:

\boxed{\text{53.3 \%}}

Explanation:

MM:           2.016      17.03

           N₂ + 3H₂ ⟶ 2NH3

m/g:            26.3  

1. Theoretical yield

(a) Moles of H₂

\text{Moles of H${_2}$} = \text{26.3 g H${_2}$} \times \dfrac{\text{1 mol H${_2}$}}{\text{2.016 g H${_2}$}} = \text{13.05 mol H${_2}$}

(b) Moles of NH₃

\text{Moles of NH${_3}$} = \text{13.05 mol H${_2}$} \times \dfrac{\text{2 mol NH${_3}$}}{\text{3 mol H${_2}$}} = \text{8.697 mol NH${_3}$}

(c) Theoretical yield of NH₃

\text{Mass of NH${_3}$} = \text{8.967 mol NH${_3}$} \times \dfrac{\text{17.03 g NH${_3}$}}{\text{1 mol NH${_3}$}} = \text{148.1 g NH${_3}$}

(d) Percent yield

\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \% = \dfrac{\text{79.0 g}}{\text{148.1 g}} \times 100 \% = \textbf{53.3 \%}\\\\\text{The percent yield is }\boxed{\textbf{53.3 \%}}

7 0
3 years ago
Read 2 more answers
Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
Korvikt [17]

Answer:

55.2kgNa_{3}AlF_{6}

Explanation:

1. First balance the equation for the synthesis of cryolite:

Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}

2. Find the limiting reagent between the Al_{2}O_{3},NaOH and HF

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}

55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

Al_{2}O_{3}:\frac{131}{1}=131

NaOH:\frac{1385}{6}=231

HF:\frac{2770}{12}=231

The Al_{2}O_{3} is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}

3 0
4 years ago
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