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3 years ago

How many significant figures should the following answer have? (23.34 cm)(90076 cm)(2.1223 cm) = cm

Chemistry

2 answers:

bezimeni [28]3 years ago

8 0

Answer : The number of significant figures are, four (4).

Explanation :

The rule apply for the multiplication and division is :

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

As we are given :

$(23.34cm)\times (90076cm)\times (2.1223cm)$

$\Rightarrow 4462000cm^3$

In this problem, the least number of significant figures is 4.

So, the correct answer is, $4.462\times 10^{6}cm^3$

Hence, the number of significant figures are, four (4).

Yuri [45]3 years ago

4 0

4 - due to being a multiplication problem the answer goes to the least amount of Sig Figs given in a previous number.

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Root nodules are associations between bacteria and plant roots responsible for

max2010maxim [7]

Answer: Root nodules are associations between bacteria and plant roots responsible for nitrogen fixation.

Explanation:

As nitrogen present in atmosphere cannot be used by plants directly.
Nitrogen present in atmosphere is converted into nitrogen compounds by bacteria which is present in root nodules of a plant.
These compounds of nitrogen are then used by plants for various functions.

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3 years ago

Convert 25 mg to unit HG

Degger [83]

Your answer would be 250,000

6 0

3 years ago

At a certain temperature the value of the equilibrium constant, Kc, is 1.27 for the reaction. 2As (s) + 3H2 (g) ⇌ 2AsH3 (g) What

Bess [88]

Answer:

$0.887$

Explanation:

Hello,

In this case, the law of mass action for the first reaction turns out:

$Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27$

Now, for the second reaction is:

$Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}$

Therefore, by applying square root for the first reaction, one obtains:

$\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}$

$\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}$

$\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}$

Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:

$Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887$

Best regards.

8 0

3 years ago

Can anyone help with this? please

Sloan [31]

Answer:

Volume of solution = 80.5 mL

Explanation:

Given data:

Molarity of solution = 4.50 mol/L

Mass of ethanol = 16.7 g

Volume of solution = ?

Solution:

Volume will be calculated from molarity formula.

Molarity = number of moles / volume in L

Number of moles:

Number of moles = mass/molar mass

Number of moles = 16.7 g/ 46.07 g/mol

Number of moles = 0.3625 mol

Volume of solution:

Molarity = number of moles / volume in L

4.50 mol/L = 0.3625 mol / volume in L

Volume in L = 0.3625 mol /4.50 mol/L

Volume in L = 0.0805 L

Volume in mL:

0.0805 L ×1000 mL/1 L

80.5 mL

3 0

3 years ago

Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of each substance ar

Nady [450]

Explanation:

Given Data:

The mass of aluminium nitrite is 72.5 g

The mass of ammonium chloride is 58.6 g

The balanced chemical equation for the reaction is given as follows.

Al(NO2)3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O

The number of moles can be determined by the formula given as follows.

Number of moles = Mass / Molar mass

The molar mass of aluminum nitrate and ammonium chloride is 164.998 g/mol and 53.49 g/mol respectively.

inserting the respective values in the formula given above.

Moles of Al(NO2)3 = 72.5 g / 164.998 g/mol = 0.439 mol

Moles of NH4Cl = 58.6 g / 53.49 g/mol = 1.096 mol

From the balanced equation,

3 moles of ammonium chloride requires 1 mole of aluminum nitrate.

So, 1 mole of ammonium chloride requires 1 / 3 mole of aluminum nitrate.

Thus, 1.096 mole of ammonium chloride will require (1 / 3) × 1.096 = 0.3653 mole of aluminum nitrite.

Here, the amount of aluminum nitrate is more than the required amount so ammonium chloride is the limiting reagent.

From the balanced chemical equation, 3 mole of ammonium chloride gives 1 mole of aluminum chloride.

So, 1.096 mole of ammonium chloride will give;

3 = 1

1.096 = x

x = (1.096 * 1 ) / 3 = 0.3653 mole of aluminum chloride.

Therefore, the number of moles of aluminum chloride is 0.3653 mol.

Since the molar mass of aluminum chloride is 133.34 g/mol

Substitute the respective values in the formula given above.

0.3653 mol = Mass / 133.34 g/mol

Mass = 0.3653 mol × 133.34 g/mol = 48.71 g

Therefore , the mass of aluminum chloride produces is 48.71 g.

The Ammonium chloride is completely used up in the reaction.

The amount of alumium nitrite used is = Number of moles * Molar mass = 0.3653 * 164.998 = 60.27

Mass of alminium nitrite left = 72.5 - 60.27 = 12.23g

8 0

3 years ago