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ki77a [65]
3 years ago
10

A teacher is preparing the materials list for an activity that all 6th graders will

Chemistry
1 answer:
Greeley [361]3 years ago
7 0

Answer:

There are 1000 mg in 1. g

There are 1000 g in 1 kg

Each students needs 2,250 mg of clay

Explanation:

In order to determine the amount of how many kilograms to order, the teacher will need to find out the total mass of clay required by the students and then the teacher will have to convert the total mass to the units required for the purchase.

The teacher would have to find out how many milligrams make one kilogram as follows

1 kg = 1000 g

1 g = 1000 mg

Therefore, 2,250 mg = 2250/1000 g = 2.25 g

2.25 g = 2.25/1000 kg = 0.00225 kg.

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What is the mass of a bar of aluminum with length 5.3cm, width 0.32 cm and height 2.34cm? The density of aluminum is 2.70g/cm3.
Tatiana [17]

Answer:

<h2>mass = 10.72 g</h2>

Explanation:

Density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

Making mass the subject we have

mass = Density \times volume

From the question

Density = 2.70 g/cm³

We assume that the aluminum is a cuboid

and volume of a cuboid is given by

Volume = length × width × height

length = 5.3 cm

width = 0.32 cm

height = 2.34 cm

Volume = 5.3 × 0.32 × 2.34 = 3.97 cm³

Substitute the values into the above and solve for the mass

We have

mass = 2.70 × 3.97

We have the final answer as

<h3>mass = 10.72 g</h3>

Hope this helps you

6 0
3 years ago
What has 20 electrons in 44 Mass
nadya68 [22]

It would be Calcium: Ca

6 0
3 years ago
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Is there an optimum amount of baking soda and vinegar to put in a plastic soda bottle rocket? How would you figure that out?
Snezhnost [94]
Add 1 tsp. of vinegar to the canister at a time, filling it almost to the top. You need to add as much vinegar to the canister as possible without the vinegar and the baking soda coming into contact when you later snap the lid onto the canister. Depending on the exact canister, this may be around 5 tsp.
3 0
3 years ago
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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --&gt; __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?
Ymorist [56]

Answer:

\% diss = 50\%

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}

Thus, it is possible to find x given the pH as shown below:

x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M

So that we can calculate the initial concentration of the acid:

\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\

[HA]_0=3.64x10^{-5}M

Therefore, the percent dissociation turns out to be:

\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

6 0
2 years ago
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