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2 years ago

How are a moon and an asteroid different? (2 points)

Chemistry

1 answer:

leonid [27]2 years ago

8 0

Answer:

Explanation:

Asteroid - A rock that rvolves arounf the sun / star.

Moon - A natural satillite that revolves around earth/ planet.

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what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

5 0

2 years ago

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

7 0

3 years ago

Amphetamine was developed initially as a substitute for a closely-related chemical derived from the Chinese herb ma huang. That

Zepler [3.9K]

Amphetamine was developed initially as a substitute for a closely-related chemical derived from the Chinese herb ma huang. That chemical is Ephedrine.

<h3>What is Ephedrine?</h3>

Ephedrine is a stimulant medication that can be considered an amine chemical compound and substituted amphetamine.

Ephedrine is used in sports to stimulate the central nervous system (C_NS) and as energy supplier.

Ep_hedrine use is prohibited and chronic usage of this substance may have problems the health.

Learn more about Ephedrine here:

brainly.com/question/1122074

4 0

1 year ago

How many protons and neutrons are in the nucleus of isotope with mass of 68.926 amu?

Sidana [21]

There are 30 protons and 39 neutrons in the nucleus.

This must me the isotope of an element with an atomic mass close to 69 u.
The only candidates are Zn and Ga.
Zn has a zinc-69 isotope with mass 68.926 u.
Ga has a gallium -69 isotope with mass 68.925 u.
The isotope is probably $_{30} ^{69}Zn$ .
It has 30 protons and 39 neutrons.

4 0

2 years ago

Two hydrogen atoms collide in a head on collision and end up with zero kinetic energy. Each then emits a photon of 121.6 nm (n=2

Zinaida [17]

Explanation:

Expression for the kinetic energy is as follows.

K.E = $\frac{1}{2}mv^{2}$