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Elenna [48]
2 years ago
13

What mass of aluminum must react with 100 g of copper (ii) phosphate? (10g)

Chemistry
1 answer:
trasher [3.6K]2 years ago
7 0
It would be 10g or 50g
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0.0145 moles of helium gas are introduced into a balloon so that the volume of the balloon is 2.54 liters. An additional amount
olga_2 [115]

Answer:

4.43L is final volume of the ballon

Explanation:

Avogadro's law of ideal gases states that <em>equal volumes of gases, at the same temperature and pressure, have the same number of molecules</em>.

The formula is:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

Where V and n are volume and moles of the gas in initial and final conditions.

If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

\frac{2.54L}{0.0145mol} =\frac{V_2}{0.0253mol}

V₂ = <em>4.43L is final volume of the ballon</em>

6 0
3 years ago
Which planets' orbits is the belt located between the main asteroid belt?
hichkok12 [17]
<span>circumstellar is the answer</span>
3 0
3 years ago
Select the three characteristics that best describe a STEM-educated student.
Lemur [1.5K]
Innovated Understanding Curious Logical
5 0
3 years ago
Read 2 more answers
139%
GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}

\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}

\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

n = m/M

\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}

% composition ofFeCl_3  

=  (9.75 / 10.39)^{*} 100

= 94%

6 0
3 years ago
How many moles are in 255 mL of a 2.35 M solution of H2SO4?
dimaraw [331]

Answer:

.59925 moles

Explanation:

3 0
3 years ago
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