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3 years ago

If the volume of a cylinder is 84,300 cubic centimeters then how many millimeters is this equal to

Chemistry

1 answer:

nevsk [136]3 years ago

8 0

Answer:

84,300,000

Explanation:

Just multiply it by 1,000. 1cm = 10mm, and a cube is the length to a sort of third degree, so you take the 10 to its third degree as well and multiply it by 1,000, instead of 10 like you would to find how many millimeters were in an amount of centimeters. If that makes sense.

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Are normal salts acidic, basic or neutral salt?

RSB [31]

Neutral salt is the answer

Hope this helps!

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3 years ago

Why can water dissolve many substances

leva [86]

Answer:

a substance that can dissolve other molecules and compounds, which are known as solutesBecause of its ability to form hydrogen bonds, , meaning that it can dissolve many different kinds of molecules.

Explanation:

6 0

3 years ago

One of the great upcoming sports in the Olympics is the sport of curling. Write a brief essay on the uses of momentum collisions

Usimov [2.4K]

The Olympic sport of curling is one that is practically designed to show Physics in motion. Curling is a sport in which two teams alternate sliding smoothed stone pucks down an ice rink court with the intent to seat their stone closest to the center of the target (called the house). Each team has eight stones, meaning that the team that goes second has the (could be) massive advantage of sending the last stone.
The mass of the stone is important in that the more massive a stone (m) and the speed at which it travels (v) dictates it's momentum (momentum=mxv). As the curling stone slides down the ice (which is relatively frictionless unless acted upon by other players or objects) and having inertia, continues in it's straight course (again, unless acted upon by outside forces). If the stone hits another stone, it transfers some of its momentum in an elastic collision to that stone and the original stone is deflected in a calculable manner.
Collisions are used in the game to either clear opponent's stones from the house or out of their defensive positions, or to make adjustments to one's stones present in the house, all based on the momentum of the moving stone, and its transference.

8 0

3 years ago

Read 2 more answers

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g

satela [25.4K]

Answer:

$5.6gNa_2SO_4$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

$n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4$

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

$m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4$

Best regards.

5 0

3 years ago

3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?

Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

$\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To balance it, we can simply add another sodium atom on the left. Hence:

$\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

$\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}$

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

$\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}$

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

$\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Given the initial value and the above ratios, this yields:

$\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Cancel like units:

$=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}$

Multiply. Hence:

$=0.3463...\text{ g H$_2$}$

Since we should have two significant values:

$=0.35\text{ g H$_2$}$

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

$\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}$

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

$\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}$

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

$\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}$

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

$\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}$

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: $\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Using the given value and the above ratios, we acquire:

$\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Cancel like units:

$\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}$

Multiply:

$\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}$

Since the resulting value should have three significant figures:

$\displaystyle = 65.2 \text{ g Al$_2$O$_3$}$

So, approximately 65.2 grams of aluminum oxide is produced.

5 0

3 years ago

Read 2 more answers