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andrey2020 [161]

2 years ago

14

Which statement is true? Postively charged objects attract other positively charged objects negatively charged objects attract o

ther negatively charged objects

1 answer:

ANEK [815]2 years ago

8 0

Answer:

Explanation:one is true because if it is positiveit would not be attracting it would be sepperating

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Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25

o-na [289]

The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation.
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72

6 0

3 years ago

Balance the chemical equation.Au plus HC l plus HNO 3 right arrow AuC l Subscript 3 Baseline plus NO plus Upper H 2 Upper OAu+HC

tino4ka555 [31]

Answer:

$Au+ 3HCl + HNO_3 \to AuCl_3+NO+2H_2O$

Explanation:

Chemical Equations are representations of chemical reactions in terms of the symbols and formulae of the elements and compounds involved. A chemical equation usually have the reactant at the left hand side while the product is on the right hand side.

A chemical Equation is of little or no value if is not in balanced equation. When an equation is balanced , the total number of atoms of any element on the left-hand side of it must be equal to the total number of atoms of that element on the right hand side.

in the given question; we are given a word problem of chemical symbol to compute and also to balance the chemical equation.

From below; the chemical equation can be written as:

$Au+ HCl + HNO_3 \to AuCl_3+NO+2H_2O$

From the above equation we will notice that it is not truly balanced ; so th balanced equation can be written as:

$Au+ 3HCl + HNO_3 \to AuCl_3+NO+2H_2O$

5 0

3 years ago

A certain indicator has a color chart that shows that when the indicator is yellow, the solution is acidic. What is true about a

bija089 [108]

It’s acidic?
It tastes sour
It has at least one hydrogen molecule

4 0

3 years ago

Please help me to do this assignment

kirill [66]

Answer:

1. Objective

2. Objective

3. Opinion

4. Objective

5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

7 0

3 years ago

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0

3 years ago