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1 year ago

HELP!! PLEASE! Describe North America’s rotation. (in terms of continental drift)

Chemistry

1 answer:

Zigmanuir [339]1 year ago

3 0

Answer:

https://www.nationalgeographic.org/encyclopedia/continental-drift/#geologic-plates-600mya

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In a 4.00 L pressure cooker, water is brought to a boil. If the final temperature is 115 °C at 3.10 atm, how many moles of steam

Olenka [21]

Answer: 0.39mol of the steam are in the cooker.

Explanation:Please see attachment for explanation

7 0

3 years ago

A sample of hydrogen gas has an initial pressure of 2.86 atm and an initial volume of 8472 mL If the volume of the

Oliga [24]

<h3>Answer:</h3>

Gas law : Boyle's law

New pressure: 66.24 atm

<h3>Explanation:</h3>

Concept tested: Gas laws (Boyle's law)

<u>We are given,</u>

Initial pressure, P₁ = 2.86 atm
Initial volume, V₁ = 8472 mL
New volume, V₂ IS 365.8 mL

We need to determine the new pressure, P₂

According to Boyle's law , the volume of a fixed mass of a gas and the pressure are inversely proportional at constant temperature.

That is, $P\alpha \frac{1}{V}$
This means , PV = k (constant)
Therefore; P₁V₁ = P₂V₂

Rearranging the formula, we can get the new pressure, P₂

P₂ = P₁V₁ ÷ V₂

= (2.86 atm × 8472 mL) ÷ 365.8 mL

= 66.24 atm

Therefore, the new pressure is 66.24 atm

5 0

2 years ago

The Hawaiian islands formed as a result of a hot spot.<br><br> A.)True <br><br> B.)False

choli [55]

Hello,

That is true.

8 0

2 years ago

Read 2 more answers

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

2 years ago

Number of moles in 31.99g O2 <br> And 176.8g NaCl

sergey [27]

#O_2

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longmapsto No\:of\:moles=\dfrac{31.99}{32}$

$\\ \sf\longmapsto No\:of\:moles\approx 1$

#NaCl

$\\ \sf\longmapsto No\:of\:moles=\dfrac{176.8}{58.5}$

$\\ \sf\longmapsto No\;of\:moles=3mol$

8 0

1 year ago