All categories

3 years ago

Evaluate 0.411 + 9.9 + 0.8 and round the answer appropriately,

Chemistry

1 answer:

Lemur [1.5K]3 years ago

8 0

Answer:

the answer is 11.111 and when you round it to the nearest one it’s 11, but to the nearest tenth it’s 10

Explanation:

You might be interested in

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

What is the ph of a 3x10-8M H+ solution

Lelu [443]

Answer:

The pH of the solution is 7, 52

Explanation:

The pH gives us an idea of the acidity or alkalinity of a solution. Is calculated:

pH = -log (H +)

pH= -log ( 3x10-8)= <em>7, 52</em>

4 0

3 years ago

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$

$75 \times \frac{1}{2} =48.2$

$34.5 \times \frac{1}{2} =72.3$

$17.25 \times \frac{1}{2} =96.4$

$8.625\times \frac{1}{2} =120.5$

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

$A=A_{0} \cdot \frac{1}{2^{n}}$

Where

$A_0$ is the initial sample amount

n = the number of half-lives that pass in a given period of time.

7 0

3 years ago

A chocolate chip cookie can be separated by physical means, and it is not the same throughout.

yarga [219]

As a heterogeneous mixture

6 0

3 years ago

Read 2 more answers

Carbon dioxide (CO2) is a gas at room temperature and pressure. However, carbon dioxide can be put under pressure to become a "s

Lunna [17]

Answer:

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

Explanation:

Step 1: Given data

The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)

The sample hasa volume of 25.0 mL

Step 2: Calculating mass of the sample

The density is the mass per amount of volume

0.469g/cm³ = 0.469g/ml

The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g

The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g

5 0

4 years ago