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hoa [83]
3 years ago
7

What is a good slogan for antimony

Chemistry
1 answer:
scZoUnD [109]3 years ago
6 0
Antimony: a phenomenal phenomenon
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True or False: As the ice warms up, the molecules get more energy and move around more from place to place to form liquid water.
Arada [10]
The answer is TRUE.

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7 0
3 years ago
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The quantity of heat required to change the temperature of 1 g of a substance by 1 c is defined as
slega [8]
I think it's calorie
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6 0
3 years ago
How many grams of Ar are there in 2.25 moles of Ar?
nydimaria [60]

Answer:

89.88 g

Explanation:

Atomic Mass of Ar: 39.948

Mass = moles * AM

Replacing moles = 2.25 and AM = 39.948 you get the mass of Ar:

Mass = 2.25 * 39.948

Mass = 89.88 g

8 0
3 years ago
What is the correct balanced equation for the reaction of sodium with water?
Wittaler [7]

Answer:

 A.    2Na       +    2H₂O   →    2NaOH   +  H₂  

Explanation:

When sodium is reacted with water, a single replacement reaction occurs. The product of the reaction is typically sodium hydroxide and hydrogen gas.

Reactants:

        Sodium +  water

Product:

        Sodium hydroxide + hydrogen gas

So;

         2Na       +    2H₂O   →    2NaOH   +  H₂  

6 0
3 years ago
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
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