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3 years ago

3. The distance from the thumb to the little finger on Mrs. McCord's hand is 98.0mm. Convert this to

Chemistry

1 answer:

lianna [129]3 years ago

4 0

Answer:

0.098m

Hope it helps

I am sure about the answer

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Where else do you see Force & Motion in real-world experiences?

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Answer:

a boy pulling a toy train, the boy is

interacting with an object while applying a force to it,

another example of contact forces is friction.

Explanation:

A contact force is when two interacting

objects

(btw love your pfp)

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2 years ago

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Solutions can be composed of:

Anika [276]

A solution may exist in any phase so your answer is D. any of the above

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3 years ago

10. Sound is characterized in how many ways?<br> A. 5<br> B. 3<br> C. 4<br> D. None of the above

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Answer:

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2 years ago

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Balance the following equation: H20 --> H2 + O2

MakcuM [25]

2H2O = 2H2 + O2.

<h3><u>Explanation</u>:</h3>

Balancing equations is very essential because of the fact that it represents the stoichiometric quantities of the reactants needed to react to form the product. The ratio of the weights of reactant and product are also very well understood from this.

Here in this equation, the water is broken into hydrogen and oxygen. The balanced reaction is

2H2O = 2H2 + O2.

Two moles of water is broken down into 2 moles of hydrogen and one mole of oxygen.

5 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

2 years ago