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2 years ago

Compare the density of the he atom with that of pure water, which is 1.0 g/cm3.

1 answer:

3 0

At room temperature, water is liquid in form. For helium, it is gas because it is the second lightest element in nature. So, you would expect the density of Helium to be much less than 1 g/cm³. Because gas is very light and occupies a lot of space, the density would result to a very small number.

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If you had 2.50 moles of anything, how many of those things would you have?

Wittaler [7]

Answer:

1.51 x 10²⁴ things

Explanation:

According to Avogadro's Constant.

1 mole of any substance, constains 6.02×10²³ particles of that subtance.

From the question,

If we have 2.50 moles of anything,

1 mole of anything ⇒ 6.02×10²³ things

2.50 moles of anything ⇒ y things

solving for y

y = (2.50× 6.02×10²³)/1

y = 15.05×10²³

y = 1.505×10²⁴

y ≈ 1.51×10²⁴

7 0

2 years ago

Use the "x is small" approximation to find the concentration of the products in the following reaction which initially contains

Mrrafil [7]

Answer:

Concentration of product at equilibrium ;

$[H^+]=0.0000229 M$

$[CN^-]=0.0000229 M$

Explanation:

$HCN(aq)\rightleftharpoons H^+(aq) + CN^-(aq)$

initially

0.85 M 0 0

(0.85-x)M x x

The equilibrium constant of reaction = $K_c= 6.17\times 10^{-10}$

The expression of an equilibrium cannot can be written as:

$K_c=\frac{[H^+][CN^-]}{[HCN]}$

$6.17\times 10^{-10}=\frac{x\times x}{(0.85-x)}$

Solving for x:

x = 0.0000229

Concentration of product at equilibrium ;

$[H^+]=0.0000229 M$

$[CN^-]=0.0000229 M$

4 0

2 years ago

The mineral stibnite, antimony(III) sulfide, is treated with hydrochloric acid to give antimony(III) chloride and hydrogen sulfi

jeka94

The balanced chemical reaction is written as:

Sb2S3 + 6HCl = 6SbCl3 + 3H2S

We are given the amount of antimony(III) sulfide to be used in the reaction. This is amount will be used for the calculations. We do as follows:

2.85 g Sb2S3 ( 1 mol / 339.715 g ) ( 6 mol SbCl3 / 1 mol Sb2S3 ) ( 228.13 g / mol ) = 11.48 g SbCl3

5 0

2 years ago

Pipettes are used when measuring the volumes of liquids with high degree of precision. The following volumes were obtained durin

never [62]

Answer:

0.075

Explanation:

First obtain the mean of the measurement;

Mean = 10.15 + 9.95 + 9.99 + 10.02/4 = 10.03

Then obtain d^2= (mean-score)^2 for each score;

(10.15-10.03)^2 = 0.0144

(9.95-10.03)^2 = 0.0064

(9.99-10.03)^2 = 0.0016

(10.02-10.03)^2= 0.0001

∑d^2= 0.0144 + 0.0064 + 0.0016 + 0.0001

∑d^2= 0.0225

Variance = ∑d^2/N = 0.0225/4 = 0.005625

Standard deviation= √0.005625

Standard deviation= 0.075

6 0

2 years ago

You are starting with 2 pounds of 30% acetic acid solution. You want to dilute

Tpy6a [65]

Answer:

The mass of water to be added is 2 pounds

Explanation:

The given parameters are;

The mass of the given solution = 2 pounds

The concentration of the given solution = 30%

The desired concentration of the solution = 15%

The mass, m of the acetic acid in the given solution = 30% × 2 pounds

m = 30/100 × 2 pounds = 0.6 pounds

To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;

15% of X = 0.6 pounds

15/100 × X = 3/20 × 0.6 pounds

∴ The mass of the solution required X = 0.6 × 20/3 = 4 pounds

The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds

Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.

8 0

2 years ago