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2 years ago

Cu + 2AgNO3 es002-1.jpg 2Ag + Cu(NO3)2 How many moles of copper must react to form 0.854 mol Ag?

1 answer:

4 0

Balance Chemical Equation is as follow,

 Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

According to Balance Equation,

2 Moles of Ag is produced by reacting = 1 Mole of Cu
So,
0.854 Moles of Ag will be produced by reacting = X Moles of Cu

Solving for X,
X = (0.854 mol × 1 mol) ÷ 2 mol

X = 0.427 Moles of Cu
Result:
0.854 Moles of Ag are produced by reacting 0.427 Moles of Cu.

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Electrons in __________ bonds remain localized between two atoms. electrons in __________ bonds can become delocalized between m

dsp73

Electrons in sigma bonds remain localized between two atoms. Sigma bond results from the formation of a molecular orbital by the end to end overlap of atomic orbitals. Electrons in pi bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals.

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2 years ago

How many molecules of CBr4 are in 250 grams of CBr4

Kazeer [188]

Answer:- $4.54*10^2^3$ molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of $CBr_4$ = 12+4(79.9) = 331.6 g per mol

let's make the set up using dimensional analysis:

$250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})$

= $4.54*10^2^3$ molecules

So, there will be $4.54*10^2^3$ molecules in 250 grams of $CBr_4$ .

8 0

1 year ago

Give the number of significant figures in this number: 40.00

Tatiana [17]

A significant figure is every symbol that made the number itself.

In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.

If you have a case with zeros in front, you take to the first non zero digits.

For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.

4 0

2 years ago

Read 2 more answers

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

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1 year ago

A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentrat

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"30.4 ppm > 4 ppm, unsafe to drink" is the one among the following choices given in the question that shows that the water should be declared unsafe for drinking. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.

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2 years ago

Read 2 more answers