A swimming pool has the dimensions of 10m, 5m, and 1m. what is the volume?

What is the force in the stretched rope that is attached to a bucket being pulled out of a well?

bezimeni [28]

The force of gravity acts upon all objects; the Earth and the object attract. There are no springs touching the object; thus, there is no spring force. A taut rope is attached to the bucket; this is the cause of the tension force.

3 0

2 years ago

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

5 0

2 years ago

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als

olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula $M_1*V_1 = M_2*V_2$ . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration Volume to be Final Concentration

added

1 M Tris 2.5 mL 10 mM

5 M NaCl 7.5 mL 150 mM

1 M Imidazole 75 mL 300 mM

$M_1*V_1 = M_2*V_2$ . is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.01 M *250mL\\V_1 = 2.5mL$

The stock concentration of NaCl (5 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.15 M *250mL\\V_1 = 7.5mL$

The stock concentration of Imidazole (1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.03 M *250mL\\V_1 = 75mL$

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

5 0

2 years ago

Consider a 0.10 M aqueous benzoic acid, CeHeCOOH. The K benzoic acid. 6.5 x 10 for A) Write a balanced equation that shows the r

7nadin3 [17]

Answer:

a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

c) pH = 2.599

Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+ + C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

7 0

2 years ago

What form of pollution can lichens be a good indicator of?

Natasha2012 [34]

Answer:

Lichens can be used as air pollution indicators, especially of the concentration of sulphur dioxide in the atmosphere. Lichens are organisms that grow in exposed places such as rocks or tree bark. They need to be very efficient at absorbing water and nutrients to grow there.

7 0

2 years ago