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2 years ago

PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST

Chemistry

2 answers:

dusya [7]2 years ago

8 0

1 to 4) it’s weird yeah doesn’t like you choose your subject

Ivahew [28]2 years ago

5 0

Answer:

up to down(1 to 6)

Explanation:

speed:

1,3,4

Velocity:

everything except 1,3,4

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2. Some nitrogen at a pressure of 35.75 p.s.i is in a 100 L container. If the container's volume is reduced to 2250 ml then what

elena-s [515]

Answer:

1455.6

Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values

35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6

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3 years ago

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gogolik [260]

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2 years ago

1500 millimeters to km

gizmo_the_mogwai [7]

0.0015 kilometers is for sure the answer!

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3 years ago

A person breathes 2.6 L of air at -11 C into her lungs, where it is warmed to 37 C. What is its new Volume?

aalyn [17]

Answer:

3.076 L.

Explanation:

The pressure is constant as it is the atmospheric pressure.

According to Charles’ law; at constant pressure, the volume of a given quantity of a gas varies directly with its temperature.

∴ V₁T₂ = V₂T₁.

V₁ = 2.6 L & T₁ = -11 °C + 273 = 262 K.

V₂ = ??? L & T₂ = 37 °C + 273 = 310 K.

∴ V₂ = (V₁T₂) / T₁ = (2.6 L)(310 K) / (262 K) = 3.076 L.

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3 years ago

I need help on both a and b of question 1

marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

$r = \frac{\Delta c}{\Delta t}$

Given the following reaction:

$2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)$

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

$r = -\frac{\Delta [N_2O_5]}{2 \Delta t}$

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

$r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}$

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

$r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s$

(b) Using the relationship derived previously, we know that:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

Rate of appearance of nitrogen dioxide is given by:

$r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}$

Which is obtained from the equation:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

If we multiply both sides by 4, that is:

$-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}$

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

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2 years ago