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stiv31 [10]
3 years ago
13

Consider four beakers labeled a, b, c, and d, each containing an aqueous solution and a solid piece of metal. identity the beake

rs in which a chemical reaction will occur and those in which no reaction will occur.
Chemistry
1 answer:
puteri [66]3 years ago
4 0
To identify in what beakers will a reaction occur, we need to know what aqueous solution is in the four beakers. Then, we also need to know what metals are placed inside those beakers. Now, we can analyze which metals might or will react to the solution in the beakers. 
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What happens to pressure if temperature increases?
krok68 [10]

as the temp increases the pressure increases to.

6 0
3 years ago
Read 2 more answers
Determine the type of reaction: AgNO3 + Cu --> Cu(NO3)2 + Ag
guajiro [1.7K]

Explanation:

Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

Cu is oxidized

Ag+ is reduced

Cu is the reducing agent

7 0
3 years ago
A solution is prepared by dissolving 91.7 g fructose in 545 g of water. Determine the mole fraction of fructose if the final vol
oee [108]

Answer:

Mole fraction = 0,0166

Explanation:

Mole fraction is defined as mole of a compound per total moles of the mixture. In the solution, the solute is fructose and the solvent is water. That means you need to find moles of fructose and moles of water.

The molecular mass of fructose is 180,16g/mol and mass of water is 18,02 g/mol. Using these values:

91,7g fructose × (1mol / 180,16g) = <em>0,509 moles of fructose</em>

545g water × (1mol / 18,02g) = <em>30,24 moles of water</em>

Thus, mole fraction of fructose is:

\frac{0,509 moles}{0,509mol + 30,24mol} = 0,0166

<em>Mole fraction = 0,0166</em>

I hope it helps!

5 0
3 years ago
What is the ph value of water and salt?
Andreyy89
The ph value of water is 7 but a salt which is  derived from a weak acid will be less than 7 and a salt which is derived from a base will be above 7.

3 0
3 years ago
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
Ket [755]

Answer:- 10 L of ethane.

Solution:- The given balanced equation is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.

Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

14LO_2(\frac{2LC_2H_6}{7LO_2})

= 4LC_2H_6

From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L

Hence, 10 L of ethane will be remaining.

5 0
3 years ago
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