All categories

2 years ago

WILL MARK BRAINLIEST

Chemistry

1 answer:

Marysya12 [62]2 years ago

8 0

Answer:

8.28 × 10⁶ N

Explanation:

Step 1: Given data

Interior surface area of the habitat (S): 110 m²

Pressure in the habitat (P): 75.3 kPa

Step 2: Convert "P" to Pascal

We will use the conversion factor 1 kPa = 10³ Pa.

75.3 kPa × (10³ Pa/1 kPa) = 75.3 × 10³ Pa

Step 3: Calculate the total force exerted (F)

We will use the following expression.

P = F/S

F = P × S

F = 75.3 × 10³ Pa × 110 m²

F = 8.28 × 10⁶ N

You might be interested in

Is it possible for the overall charge of an ionic compound to be -1. true false

AnnZ [28]

Answer:

The correct answer is False

8 0

2 years ago

How might you tell if a food contains an acid as one of its ingredients?

vladimir2022 [97]

We can collect a sample of that food and add a drop of blue litmus solution. If the color of the litmus solution changes to red, it contains an acid as one of its ingredients.

7 0

3 years ago

Read 2 more answers

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is

lianna [129]

Answer:

21.6 g

Explanation:

The reaction that takes place is:

CH₄ + 2O₂ → CO₂ + 2H₂O

First we convert the given masses of both reactants into moles, using their respective molar masses:

9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄

64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we calculate how many moles of water are produced, using the number of moles of the limiting reactant:

0.6 mol CH₄ * $\frac{2molH_2O}{1molCH_4}$ = 1.2 mol H₂O

Finally we convert 1.2 moles of water into grams, using its molar mass:

1.2 mol * 18 g/mol = 21.6 g

4 0

2 years ago

You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.

8 0

2 years ago

26.6 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 16.0 mL sample of the resulting solution is then diluted to

frozen [14]

Answer:

In the final solution, the concentration of sucrose is 0.126 M

Explanation:

Hi there!

The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:

Ci · Vi = Cf · Vf

Where:

Ci = concentration of the original solution

Vi = volume of the solution taken to prepare the more diluted solution.

Cf = concentration of the more diluted solution.

Vf = volume of the more diluted solution.

For the first dillution:

26.6 ml · 2.50 M = 50.0 ml · Cf

Cf = 26.6 ml · 2.50 M / 50.0 ml

Cf = 1.33 M

For the second dilution:

16.0 ml · 1.33 M = 45.0 ml · Cf

Cf = 16.0 ml · 1.33 M / 45.0 ml

Cf = 0.473 M

For the third dilution:

20.0 ml · 0.473 M = 75.0 ml · Cf

Cf = 20.0 ml · 0.473 M / 75.0 ml

Cf = 0.126 M

In the final solution, the concentration of sucrose is 0.126 M

7 0

2 years ago