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Andrews [41]
3 years ago
10

In the pictured reaction, NH4, would be acting as the

Chemistry
1 answer:
Svetach [21]3 years ago
7 0

Answer:

conjugate acid

Explanation:

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I would cry pls answer with explanation.​
lord [1]

Answer:

1) Test tube 3, and 1.

2) Test tube 4, and 2.

3) Test tube 3.

4) Test tube 1.

Explanation:

Test tubes 4, and 2 don't react because the sodium carbonate (NaCO3) needs to become sodium hydrogen carbon (NaHCO3) to form a possible reaction. NaCO3 + C6H8O7 → No reaction.

NaHCO3 + C6H8O7 → NaC6H7O7 + H2O + CO2

5 0
3 years ago
Dalton’s completing an investigation in the science lab. He observes that a sample of liquid turns to gas at 135°C. What’s this
marishachu [46]
Boiling point.

It is the temperature at which the vapor pressure of the liquid equals the vapor pressure of tha atmosphere that surrounds it, which permit to defeat the attraction among the molecules in the liquid phase and pass to the gas phase.
3 0
3 years ago
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Show
SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}     .....(1)

Molarity of HCl solution = 0.75 M

Volume of HCl solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles  

CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)  

According to stoichiometry :

2 moles of HCl require = 1 mole of CaCO_3

Thus 0.01875 moles of HCl will require=\frac{1}{2}\times 0.01875=0.009375moles  of CaCO_3

Mass of CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g

Thus 0.9375 g of CaCO_3 is required to react with 25.0 ml of 0.75 M HCl

6 0
3 years ago
Please help ASAP!! I’m stuck!!
Gnoma [55]

Answer:

The answer of this question is 3

8 0
3 years ago
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Gas particles do not have much energy of motion.<br><br> True or False?
QveST [7]
False is the answer
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4 years ago
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