All categories

3 years ago

HELP! ONLY HAVE & MINUTES LEFT!

Chemistry

1 answer:

Ainat [17]3 years ago

3 0

All the objects are formed from the gas and dust orbitting the sun

You might be interested in

A perfect cube of aluminum metal was found to weigh 20.00 g. The density of aluminum is 2.7 g/ml. What are the dimensions of the

denis23 [38]

Answer:

Height = 1.9493 cm

Width = 1.9493 cm

Depth = 1.9493 cm

Solution:

Data Given:

Mass = 20 g

Density = 2.7 g/mL

Step 1: Calculate the Volume,

As,

Density = Mass ÷ Volume

Or,

Volume = Mass ÷ Density

Putting values,

Volume = 20 g ÷ 2.7 g/mL

Volume = 7.407 mL or 7.407 cm³

Step 2: Calculate Dimensions of the Cube:

As we know,

Volume = length × width × depth

So, we will take the cube root of 7.407 cm³ which is 1.9493 cm.

Hence,

Volume = 1.9493 cm × 1.9493 cm × 1.9493 cm

Volume = 7.407 cm³

4 0

3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

3 years ago

Which radioactive isotope would take the least amount of time to become stable? rubidium-91 iodine-131 cesium-135 uranium-238

Masteriza [31]

The answer is rubidium -91 because it takes a shorter time of 58.4 seconds to become stable.

5 0

3 years ago

Read 2 more answers

What would happen to other populations if they didn't have limitng factors?

Montano1993 [528]

Answer:

In the natural world, limiting factors like the availability of food, water, shelter and space can change animal and plant populations. Other limiting factors, like competition for resources, predation and disease can also impact populations. Other changes in limiting factors will cause a population to decrease.

8 0

3 years ago

The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what

valina [46]

As the atmospheric pressure is, P = dgh

Here d is the density of the mercury,

g is gravitation = 9.8 m/s²

h is height of the column, P = 751 torr = (751 torr × 1 atm / 760 torr) (101325 Pa) (1 N/m² / 1 Pa) = 100125 N/m²

Where, 1 N = 1 Kg / ms²

Thus, P = 100125 Kg / m³. s²

Therefore, height of the mercury column, when the atmospheric pressure is 751 torr,

h = P / gd

= (100125 kg / m³. s²) / (9.8 m/s²) (13.6 × 10³ kg / m³) = 0.751 m

As, d₁h₁ = d₂h₂

Here, d₁ is the density of the non-volatile liquid = 1.20 g/ml

d₂ is the density of the mercury = 13.6 g/ml

h₂ = 0.751 m

Thus, putting the values we get,

h₁ = d₂h₂ /d₁ = 13.6 g/ml × 0.751 m / 1.20 g/ml

= 8.5 m

3 0

3 years ago