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3 years ago

Compare the the volumes and densities of two pieces of lead: one with a mass of 25 g and the other with a mass of 75 g.

Chemistry

1 answer:

enot [183]3 years ago

4 0

Answer:

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Explanation:

bklgffmgd. but the gdfb by the first gunnk

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determine the maximum amount of NaN03 that was produced during the experiment. Explain how you determined the amount

Angelina_Jolie [31]

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl is the limiting reactant and 9 moles of NaNO3 is obtained.

7 0

2 years ago

tankabanditka [31]

Watershed
A watershed is an entire river system of an area drained by a river and its tributaries

3 0

3 years ago

A hydrogen bond is a dipole-dipole attraction between a partially positive hydrogen atom and the unshared electron pair of a str

podryga [215]

True! The answer is true because I did this today and got that answer and it was right!

6 0

2 years ago

Two cylinders are made of the same material. Cylinder A is one-fourth (1/4) the length of cylinder B and it has a radius that is

tankabanditka [31]

Answer:

$M_{A}$ : $M_{B}$ = 4 : 1

Explanation:

Given that:

Volume of a cylinder = $\pi$ $r^{2}$ h

For cylinder A, $l_{A}$ = $h_{A}$ = $\frac{1}{4} l_{B}$ and $r_{A}$ = 4 $r_{B}$ .

Volume of cylinder A = $\pi$ $(4r_{B}) ^{2}$ × $\frac{1}{4} l_{B}$

= 4 $\pi$ $r_{B} ^{2}$ $l_{B}$

Volume of cylinder B = $\pi$ $(r_{B}) ^{2}$ $l_{B}$

= $\pi$ $r_{B} ^{2}$ $l_{B}$

To determine the ratio of their masses, density (ρ) is defined as the ratio of the mass (M) of a substance to its volume (V).

i.e ρ = $\frac{M}{V}$

Thus, since the cylinders are made from the same material, they have the same density (ρ). So that;

density of A = density of B

density of A = $\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }$

density of B = $\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }$

⇒ $\frac{M_{A} }{4\pi r_{B} ^{2}l_{B} }$ = $\frac{M_{B} }{\pi r_{B} ^{2}l_{B} }$

The ratio of mass of cylinder A to that of B is given as;

$\frac{M_{A} }{M_{B} }$ = $\frac{4\pi r_{B} ^{2} l_{B} }{\pi r_{B} ^{2} l_{B} }$

⇒ $\frac{M_{A} }{M_{B} }$ = $\frac{4}{1}$

Therefore, $M_{A}$ : $M_{B}$ = 4 : 1

8 0

3 years ago

Read 2 more answers

Sodium tends lose a single electron in natural settings. Based on what you know, what are two other elements that tend to do the

Oksi-84 [34.3K]

Answer:

Lithium and Sodium

Explanation:

Losing and electron in natural setting is characterizes of elements in group one. These are elements known as the alkaline earth metals. They are the most electropositive elements on the periodic table.

These elements ionize by losing an electron yin their outermost shell to attain the configuration of the nearest noble gas. These elements are usually found in combined and rarely seen in uncombined state principally due to their very reactive nature.

Sodium naturally would ionize by losing one electron. Other elements capable of this even at a better rate because they are more electropositive are potassium and lithium. Both are also group one alkaline metals

6 0

3 years ago