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3 years ago

How does heat gases cause increase in pressure?

Chemistry

2 answers:

weqwewe [10]3 years ago

6 0

Answer:

I put this so the guy up top can have brainliest

Explanation:

your welcome

Amanda [17]3 years ago

5 0

Answer:

Explanation:

A is the correct answer.

If you found my answer useful then please mark me brainliest

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What is the mass of ether(0.71) which can be put into a beaker holding 130ml

UkoKoshka [18]

D=m/v ⇒ m=d*v

d=density
m=mass
v=volume

d(ether)=0.71 gr/cm³=0.71 gr/ ml
v=130 ml

m=d*v
m=0.71 gr/ml*(130 ml)=92.3 g

Solution: m=92.3 g

6 0

3 years ago

Since both HBr and KOH are strong, we expect these reactions to occur fully. You mix 2 moles of HBr with 3 moles of KOH in enoug

AysviL [449]

Answer: pH = 14

Explanation: Please see the attachments below

8 0

3 years ago

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.

6 0

3 years ago

What is atom economy?

bonufazy [111]

Answer:

The conversion efficiency of a chemical process.

Explanation:

Hope this helps!

5 0

3 years ago

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Romashka-Z-Leto [24]

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we want to find $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ?$ when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate this equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Plug known values into the equation and solve for dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Divide both sides by 100π to solve for dh/dt.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0

2 years ago