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3 years ago

Write the NET IONIC EQUATION between zinc and hydrochloric acid.

Chemistry

1 answer:

makkiz [27]3 years ago

6 0

Answer:

$Zn_{(s)}+2H^{+}_{(aq)}\rightarrow Zn^{2+}_{(aq)}+H_2_{(g)}$

Explanation:

Zinc reacts with hydrochloric acid to given zinc chloride and hydrogen gas. The liberation of the hydrogen gas can be confirmed by burning the gas. If the gas burns with pop sound, it means it is hydrogen gas

The molecular chemical equation for the reaction is:

$Zn_{(s)}+2HCl_{(aq)}\rightarrow ZnCl_2_{(aq)}+H_2_{(g)}$

The total ionic equation is:

$Zn_{(s)}+2H^{+}_{(aq)}+2Cl^{-}_{(aq)}\rightarrow Zn^{2+}_{(aq)}+2Cl^{-}_{(aq)}+H_2_{(g)}$

The net ionic equation is:

$Zn_{(s)}+2H^{+}_{(aq)}\rightarrow Zn^{2+}_{(aq)}+H_2_{(g)}$

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Two samples of sodium chloride with different masses were decomposed into their constituent elements. One sample produced 1.731.

Amanda [17]

Answer:

The answer to your question is letter c) 6.09 g of sodium and 9.38 g of chlorine.

Explanation:

This problem is solve using rule of three

We know that the proportion Sodium to Chloride is 1 to 1 in sodium chloride, so we have to look for this proportion in the options

AM Sodium = 23 g

AM Chlorine = 35.5 g

Sodium Chlorine

23 g ---------------- 1 mol 35.5 g -------------- 1 mol

1713.73g ------------- x 2666.6 g ------------- x

x = 1713.73/23 = 74.51 x = 2666.6/35.5 = 75.12

These values are very similar, we have to look for the proportion in the options

a) 6.09g of sodium = 0.26 mol

4,87 g of chlorine = 0.14 mol These numbers are not very similar

b) We have 0.26 mol of Na

0.037 mol of Cl This is not the answer

c) We have 0.26 mol of Na

0.26 mol of Cl These numbers are the same, the proportion is 1:1, this is the answer

d) We have 0.26 mol of Na

0.36 mol of Cl This is not the answer

8 0

3 years ago

Help please I’ll mark as brainiliest

fredd [130]

Answer:

B B

W BW BW

Explanation:

Hope this helps

8 0

3 years ago

C12H22011+1202-->12CO2+11H20

kogti [31]

Answer:

0.185moles

Explanation:

Given parameters:

Volume of O₂ = 49.8L

Unknown:

Number of moles of sucrose required = ?

Solution:

We can assume that the reaction takes place at standard temperature and pressure.

From this, we can find the number of moles of oxygen that reacted and extrapolate to that of sucrose.

Chemical equation;

C₁₂H₂₂0₁₁ + 120₂ → 12CO₂ + 11H₂0

Number moles = $\frac{volume of gas}{22.4}$ at STP

Number of moles of oxygen gas = $\frac{49.8}{22.4}$ = 2.22moles

12 moles of oxygen gas combines with 1 mole of sucrose

2.22 moles of oxygen gas will combine with $\frac{2.22}{12}$ = 0.185moles

4 0

4 years ago

The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0

3 years ago

A student makes the following two claims about chemical reactions:

Svet_ta [14]

I would agree with the second one, not the first. You can't always see the chemical reaction, and it isn't always sudden. But the second claim is true.

7 0

4 years ago