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3 years ago

What elements makeup Mars' atmosphere?

Chemistry

1 answer:

Genrish500 [490]3 years ago

8 0

Answer:

Carbon dioxide or co^2

Explanation:

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Glucose is absorbed into the blood in the small intestine. Most of the glucose is absorbed by diffusion. How does the glucose co

Sonbull [250]

Answer: The concentration in the blood is lower

Explanation:

4 0

3 years ago

Read 2 more answers

How many molecules of O2 are present in a 3.90L flask at a temperature of 273K and a pressure of 1.00atm?

Lisa [10]

Based on the ideal gas relation:

PV = nRT

where P = pressure ; V = volume ; T = temperature

n = number of moles; R = gas constant = 0.0821 L atm/mol-K

Step 1: Find the number of moles of O2

n = PV/RT = 1 * 3.90/0.0821*273 = 0.1740 moles

Step 2: Calculate the molecules of O2

Now, 1 mole of O2 corresponds to 6.023 * 10²³ molecules of O2

Therefore, 0.1740 moles of O2 corresponds to-

0.1740 moles of O2 * 6.023*10²³ molecules of O2/1 mole of O2

= 1.048 * 10²³ molecules of O2

4 0

3 years ago

Read 2 more answers

Which two substances are reactants in the following chemical equation?

krok68 [10]

Both A and D
Because these compunds of reactant and the compounds of the product . The reactant goint to equation

4 0

2 years ago

Please help me understand how to...

KonstantinChe [14]

Answer:

$m_{Cr_2O_3}^{actual}=62.4gCr_2O_3$

Explanation:

Hello!

In this case, according to the reaction:

$4Cr+3O_2\rightarrow 2Cr_2O_3$

We can see there is a 4:2 mole ratio between chromium and chromium (III) oxide, this, for the given 56.2 g of chromium, the theoretical yield of the oxide product is computed down below:

$m_{Cr_2O_3}^{theoretical}=56.2gCr*\frac{1molCr}{52.0gCr}*\frac{2molCr_2O_3}{4molCr} *\frac{151.99gCr_2O_3}{1molCr_2O_3} =82.13gCr_2O_3$

Now, considering the 76.0-% yield for this reaction, the actual yield turns out:

$m_{Cr_2O_3}^{actual}=82.13gCr_2O_3*\frac{76.0gCr_2O_3}{100gCr_2O_3} \\\\m_{Cr_2O_3}^{actual}=62.4gCr_2O_3$

Best regards!

3 0

3 years ago

Mercury has a specific gravity of 13.6. how many milliliters of mercury have a mass of 0.35 kg

SashulF [63]

The formula for specific gravity is:

$Specific gravity = \frac{\rho _{substance}}{\rho _{water}}$

where $\rho _{substance}$ is the density of the substance and $\rho _{water}$ is the density of water.

The density of water, $\rho _{water}$ = $1 g/mL$

Substituting the values in above formula we get,

$13.6 = \frac{\rho _{substance}}{1}$

$\rho _{substance} = 13.6 g/mL$

The formula of density is:

$density = \frac{mass}{volume}$

The density of mercury is $13.6 g/mL$

The mass of mercury is $0.35 kg = 0.35 kg \times 1000 \frac{g}{kg} = 350 g$

Substituting the values in density formula:

$13.6 g/mL = \frac{350 g}{volume}$

$volume = \frac{350 g}{13.6 g/mL} = 25.73 mL$

6 0

3 years ago