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3 years ago

Whats Potassium bond type? Please help!

Chemistry

2 answers:

olga_2 [115]3 years ago

5 0

Answer:

A metallic bond.

Explanation:

Potassium is a metal (alkali metal), hence its bonds are metallic bonds.

Hope this helped!

Valentin [98]3 years ago

3 0

Answer:

metallic bond............

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The system brings molecules to the cells. It includes blood vessels and the heart.

makkiz [27]

I think it’s the circulatory system. it delivers oxygen and nutrients to the cells and pumps blood to the heart. hope this helps:)

3 0

2 years ago

a radiator is filled with a mixture of 3.25kg ethylene glycol (C2H6O2) in 7.75kg of water. calculate the molality of this questi

Vladimir [108]

3.25 kg in g = 3.25 * 1000 = 3250 g

Molar mass C₂H₆O₂ = 62.0 g/mol

Mass solvent = 7.75 kg

Number of moles:

n = mass solute / molar mass

n = 3250 / 62.0

n = 52.419 moles

Molality = moles of solute / kilograms of solvent

M = 52.419 / 7.75

M = 6.7637 mol/kg

hope this helps!

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

The tarnish that forms on objects made of silver is solid silver sulphide. This can be removed by reacting it with aluminium met

steposvetlana [31]

Answer:

5.83 mol.

Explanation:

From the balanced reaction:

2Al + 3Ag₂S → 6Ag + Al₂S₃,

It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.

Al reacts with Ag₂S with (2: 3) molar ratio.

So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.

The limiting reactant is Ag₂S.

The excess "left over" reactant is Al.

The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.

6 0

3 years ago

Put 30,000,000,000 into scientific notation

Neko [114]

Your number in decimal form is
3,000,000
.
To get to standard scientific notation, we move the decimal point so there is only one non-zero digit in front of the decimal point.
So,
3,000,000
becomes
3.000,000
.
The trailing zeroes are not significant, so
3.000,000
becomes
3
.
We moved the decimal point six places, so the exponent is
6
.
We moved the decimal point to the left, so the exponent is positive.
The exponential part is therefore
10
6
.
3,000,000=3 X10^6

8 0

3 years ago

Read 2 more answers