All categories

3 years ago

4. How do these measurements compare? 2500 meters and 200 grams *

Chemistry

1 answer:

Alekssandra [29.7K]3 years ago

8 0

Answer:

See Explanation

Explanation:

Given

2500 meters

200 grams

Required

How do they compare?

You can't compare these two measurements in terms or size, units and measurements because they measure two different quantities

2500 meters measures length or distance

while

200 grams measures mass of objects

You might be interested in

Use your periodic table to determine which element in the group below would be least likely to react with the other elements ?

kipiarov [429]

Answer:

Neon

Explanation:

Neon is a noble gas. Noble gases do not react with other elements. Carbon, Boron, and Oxygen would all react with each other.

8 0

3 years ago

Read 2 more answers

Describe the relationship between an enzyme, substrate, and active site.

alexdok [17]

So what I know is that enzyme and substrate are like lock and key meaning that when the active site of the enzyme changes, the enzyme will not fit to the substrate which will lead the enzyme to denature. Hope this helps.

8 0

3 years ago

The specific gravity of gold is 19.3. What is the length of one side of a 0.4 kilograms cube of solid gold, in units of inches?

zepelin [54]

Specif gravity = density of the material / density of water

density of the material = specific gravity * density of water

density of gold = 19.3 * 1 g/mL = 19.3 g/mL

density = mass / volume ==> Volume = mass / density

Volume = 0.4 kg * 1000 g/kg / 19.3 g/mL = 20.725 mL

Length of one side of the cube = $\sqrt[3]{20.725 {cm}^{3} }$ = 2.75 cm

Pass cm to inches
2.75 cm * 1 inch / 2.54 cm = 1.08 inch

8 0

3 years ago

Read 2 more answers

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

What is a hypothesis?

gladu [14]

Answer: a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.

Explanation: definition

3 0

3 years ago