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2 years ago

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Chemistry

1 answer:

alukav5142 [94]2 years ago

5 0

decameters - meters: multiply by 10

meters to meters: multiply by 1

centimeters to meters: divide by 100

millimeters to meters: divide by 1000

For the rows at the bottom:

hectometer row: 100, multiply by 100, 4500

decameter row: 10, multiply by 10, 450

meter row: 1, multiply by 1, 45

decimeter row: 0.1, divide by 10, 4.5

centimeter row: 0.01, divide by 100, 0.45

im guessing theres a millimeter row at the bottom:

millimeter row: 0.001, divide by 1000, 0.045

hope this helps!

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Consider the reaction ch3oh(l)→ch4(g)+1/2o2(g). part a calculate δrh∘298.

Vsevolod [243]

To determine the standard heat of reaction, ΔHrxn°, let's apply the Hess' Law.

ΔHrxn° = ∑(ν×ΔHf° of products) - ∑(ν×ΔHf° of reactants)
where
ν si the stoichiometric coefficient of the substances in the reaction
ΔHf° is the standard heat of formation

The ΔHf° for the substances are the following:
CH₃OH(l) = -238.4 kJ/mol
CH₄(g) = -74.7 kJ/mol
O₂(g) = 0 kJ/mol

ΔHrxn° = (1 mol×-74.7 kJ/mol) - ∑(1 mol×-238.4 kJ/mol)
ΔHrxn° = +163.7 kJ

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2 years ago

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He standard reduction potentials of lithium metal and chlorine gas are as follows: reaction reduction potential (v) li+(aq)+e−→l

AURORKA [14]

2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)

The cell potential of the reaction above is +4.40V

<em><u>calculation</u></em>

Cell potential =∈° red - ∈° oxidation

in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04

Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V

cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V

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3 years ago

What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?

lisabon 2012 [21]

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

$2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3$ ;

$4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}$ .

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

$3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}$ .

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

$2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}$

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

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3 years ago

Na +H₂O- NaOH +H₂ balance

Whitepunk [10]

Hey there!

Na + H₂O → NaOH + H₂

First, balance O.

One on the left, one on the right. Already balanced.

Next, balance H.

Two on the left, three on the right. Let's add a coefficient of 2 in front of NaOH and a coefficient of 2 in front of H₂O, so we have 4 on each side.

Na + 2H₂O → 2NaOH + H₂

Lastly, balance Na.

One on the left, two on the right. Add a coefficient of 2 in front of Na.

2Na + 2H₂O → 2NaOH + H₂

This is our final balanced equation.

Hope this helps!

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3 years ago