A ______ is a hole where the steam looks a lot like a smoke

Humans exhale carbon dioxide when they breathe. If the concentration of carbon dioxide in the air that

Verizon [17]

Answer:

4 people

Explanation:

In order to know this, we need to gather the data.

First we have 3.5x10⁴ g of LiOH. This grams are to be last for the whole 9 days mission shuttle.

So, we first need to know how much of CO₂ will be neutralized by this mass. This can be done calculating the moles of LiOH, and then, the moles of CO₂

The molecular weight of LiOH is 23.94 g/mol and for CO₂ is 44 g/mol so:

CO₂ + 2LiOH -------> Li₂CO₃ + H₂O

moles of LiOH = 3.5x10⁴ / 23.94 = 1461.988 moles

We have a mole ratio of 1:2 so the moles of CO₂:

moles CO₂ = 1461.988 / 2 = 730.994 moles

The mass of CO₂ for the 9 days will be:

m CO₂ = 730.994 * 44 = 32163.74 g

Now, let's see how much it has to be used per day:

m CO₂/d = 32163.74 / 9 = 3573.75 g per day

This means that we can only take a few astronauts, and this number is:

n° astronaut = 3573.75 / 8.8x10²

<h2>n° astronaut = 4.06</h2><h2>In other words, only 4 astronaut</h2>

3 0

3 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

<u>1. Table</u>

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

<u>2. Orders</u>

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

<u>3. Rate equation</u>

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

<u>4. Rate constant, k</u>

<u />

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

$rate=k[A]^2C=0.0075[A]^2[C]$

<u>6. Substitute</u>

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

7 0

3 years ago

Does the entropy (△S) increase or decrease in each of the following processes? The wind blows a pile of leaves all over the yard

garik1379 [7]

B, increasing decreasing is my opinion

6 0

3 years ago

What is the concentration of a saturated solution of SrSO4? SrSO4 has a Ksp of 3.2 x 10–7

Andrej [43]

SrSo4 = Sr(2+) + SO4(2-)

Let’s say that the initial concentration of SrSo4 was 1. ( or we have 1 mole of this reagent).

When The reaction occurs part of SrSo4is dissociated. And we get X mole Sr(2+) and So4(2-).
Ksp=[Sr(2+)]*[SO4(2-)]
X^2=3.2*10^-7
X=5.6*10^-4

5 0

3 years ago

According to the vsepr model the arrangement of electron pairs around nh3 and ch4 is

bazaltina [42]

The arrangement of electron pairs around CH4 and NH3, According to the VSEPR model is the same, because in each case there are the same number of electron pairs around the central atom. So the NH3 and CH4 arrangement of electron pairs is the same because in each case there are the same number of electron pairs around the central atom.

7 0

3 years ago

A ______ is a hole where the steam looks a lot like a smoke​

A ______ is a hole where the steam looks a lot like a smoke