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Wittaler [7]
3 years ago
14

ZnCI2 + H2SO4 = ZnSO4 + HCI

Chemistry
2 answers:
Alex Ar [27]3 years ago
8 0

Answer:

it makes no sense

Explanation:

Nimfa-mama [501]3 years ago
4 0

Answer:

ZnCl2 + H2SO4 = ZnSO4 + 2HCl

Explanation:

I hope it's helpful!

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hey guys, i found this marble outside and covered in grease. so i took it in to wash it. can you please tell me what it is made
anzhelika [568]
Most likely metal but possibly steel
5 0
3 years ago
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A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound
Karolina [17]

Answer:

The empirical formula of the compound is C_{0.504}HO_{1.008}.

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen (H) is scaled up to a mole, since it has the molar mass, and both carbon (C) and oxygen (O) are also scaled up in the same magnitude. The empirical formula is of the form:

C_{x}HO_{y}

Where x, y are the number of moles of the carbon and oxygen, respectively.

The scale factor (r), no unit, is calculated by the following formula:

r = \frac{M_{H}}{m_{H}} (1)

Where:

m_{H} - Mass of hydrogen, in grams.

M_{H} - Molar mass of hydrogen, in grams per mole.

If we know that  M_{H} = 1.008\,\frac{g}{mol} and m_{H} = 0.2\,g, then the scale factor is:

r = \frac{1.008}{0.2}

r = 5.04

The molar masses of carbon (M_{C}) and oxygen (M_{O}) are 12.011\,\frac{g}{mol} and 15.999\,\frac{g}{mol}, then, the respective numbers of moles are: (r = 5.04, m_{C} = 1.2\,g, m_{O} = 3.2\,g)

Carbon

n_{C} = \frac{r\cdot m_{C}}{M_{C}} (2)

n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }

n_{C} = 0.504\,moles

Oxygen

n_{O} = \frac{r\cdot m_{O}}{M_{O}} (3)

n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }

n_{O} = 1.008\,moles

Hence, the empirical formula of the compound is C_{0.504}HO_{1.008}.

3 0
3 years ago
What is the function of hemoglobin in the body?
aleksley [76]
Main function of haemoglobin in the body is to transport oxygen to every cell/organ of the body

Hope this helps!! 
7 0
4 years ago
Read 2 more answers
What is the pH of 0.0050 HF (Ka = 6.8 x 10-4)? <br> 2.73 <br> 11.70 <br> 11.27 <br> 2.30
alexgriva [62]

The correct answer is 2.73.

HF is a weak acid which partially dissociates to release H+ and F-

                HF           →         H⁺   + F⁻

Initial      0.0050                   0       0

Change    -x                        +x       +x

Equilibrium 0.0050–x         +x     +x

Solve by using the equilibrium expression: =  [H⁺] [F⁻]/ [HF]

6 .8 x 10⁻⁴= x. x / 0.0050   –x

6 .8 x 10⁻⁴= x² /0.0050  

x²  = 6 .8 x 10⁻⁴ x 0.0050  

x²  = 3.4 x 10⁻⁶

x = 3.4 x 10⁻⁶

[H⁺]  = 1.84 x 10⁻³

pH = - log [H⁺] = - log (1.84 x 10⁻³)

pH = 2.73

7 0
3 years ago
Ionic bonds result from high electronegativity differences.<br> a. True<br> b. False
nata0808 [166]
True; i<span>onic bonds result from high electronegativity differences</span>
3 0
3 years ago
Read 2 more answers
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