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3 years ago

Will a horse galloping in a field produce and average velocity of zero

Chemistry

1 answer:

motikmotik3 years ago

5 0

Answer:

Explanation:

Depending on the mass of the horse and the speed, velocity will change.

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What is the pOH of a<br> 5.6 x 10-5 M solution of cesium<br> hydroxide (CsOH)?

Rudiy27

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

7 0

3 years ago

12 g of powdered magnesium oxide reacts with nitric acid to

galben [10]

Answer:

80.8 g

Explanation:

First, let's write a balanced equation of this reaction

MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O

Now let's convert grams to moles

We gotta find the weight of MgO

24 + 16 = 40 g/mol

12/40 = 0.3 moles of MgO

We can use this to find out how much Magnesium Nitrate will be formed

0.3 x 1 MgO / 1 Mg(NO₃)₂ = 0.3 moles of Magnesium Nitrate formed

Convert moles to grams

Find the weight of Mg(NO₃)₂ but don't forget that 2 subscript acts as a multiplier of whatever is inside that parenthesis.

24 + 14 x 2 + 16 x 3 x 2 = 148 g/mol

148 x 0.3 = 80.8 g

4 0

3 years ago

What domain is an onion

Nana76 [90]

The domain that is an onion is the <span>Eukaryote.

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3 0

3 years ago

Read 2 more answers

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

IceJOKER [234]

Answer:

$1.75\cdot 10^{-4} M$

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

$S = k_H p^o$

Where $k_H$ is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

$S_1 = 2.67\cdot 10^{-4} M$

Also, at sea level, we have an atmospheric pressure of:

$p = 1.00 atm$

Given mole fraction:

$\chi_{O_2} = 0.209$

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

8 0

3 years ago

How many kJ are 3,340J?

forsale [732]

The answer I believe is 3.340kj.

8 0

3 years ago