How to calculate how muchyou would get from the lottery?

Molodets [167]

Answer:

do it yourself

mate.

<h2>because doing work yourself and understanding increases your knowledge understand</h2>

<h3>and mark my answer as Brainlist because marking my answer as Brainlist you would not be losing anything and about your answer do yourself</h3>

7 0

3 years ago

A piece of metal with a mass of 159 g is placed in a 50 mL graduated cylinder. The water level rises from 20. mL to 41 mL. What

almond37 [142]

Volume of water displaced is equal to the volume of the metal. So 41 - 20 = 21 ml. Density = mass / volume so density = 159 / 21 = 7.57 g/ml

8 0

3 years ago

A 25 kg rock is placed in a graduated cylinder with water.the volume of the fluid is 18.3ml.calculate the density of the rock in

ipn [44]

Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m) = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

$p = \frac{m}{v}$

First let's convert 25 Kg to g;

1 Kg = 1000 g

25 Kg = ?

$= \frac{25 × 1000}{1}$

= 25000 g

Substitute the values into the formula:

$p = \frac{25000 g}{18.3 ml}$

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is 1366.120 g/mL.

8 0

2 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

Read 2 more answers

1. How many protons does 14/6 C contain? What element is this?

Harman [31]

Answer:

It has 6 protons and its Carbon 14

Explanation:

3 0

3 years ago