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Tasya [4]
3 years ago
5

How do I convert/scale a recipe using a t chart

Chemistry
1 answer:
Rina8888 [55]3 years ago
6 0

Answer:

Obtain the conversion factor by dividing the required yield (from Step 2) by the old yield (from Step 1). That is, conversion factor = (required yield)/(recipe yield) or conversion factor = what you NEED ÷ what you HAVE.

Explanation:

You might be interested in
a student determined that it requires 106220 j of energy to vaporize 47g of water. is the student is right​
shusha [124]
<h2>Answer:</h2>

He is right that the energy of vaporization of 47 g of water s 106222 j.

<h3>Explanation:</h3>

Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.

In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.

It means for vaporizing 18 g, 40.65 kJ energy is needed.

So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ

Hence the student is right about the energy of vaporization of 47 g of water.

7 0
3 years ago
What is the volume of 8.80 g of CH4 gas at STP?
Ann [662]

Answer:

12.32 L.

Explanation:

The following data were obtained from the question:

Mass of CH4 = 8.80 g

Volume of CH4 =?

Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:

Mass of CH4 = 8.80 g

Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol

Mole of CH4 =?

Mole = mass/Molar mass

Mole of CH4 = 8.80 / 16

Mole of CH4 = 0.55 mole.

Finally, we shall determine the volume of the gas at stp as illustrated below:

1 mole of a gas occupies 22.4 L at stp.

Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.

Thus, 8.80 g of CH4 occupies 12.32 L at STP.

6 0
3 years ago
Metallic elements can be recovered from ores that are oxides, carbonates, halides, or sulfides. Give an example of each type.
vredina [299]

Metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

Ores which is deposited in earth's crust which contain minerals and metals. Metals can be obtained economically and sold commercially.

<h3>How metals obtained from sulphide or carbonate ore? </h3>

As we get to know that it is easy to obtain metals from their oxides. So, firstly ores which is found in the form of carbonates and sulphide are converted into their oxides by using the process of calcination and roasting.

The metals which is in the middle of the activity series are moderately reactive. These metals are found in the crust of the earth is mainly found as oxides, sulphides, or carbonates.

A metal which can occur in the form of sulphide ore is lead.

A metal which can occur in the form of oxide ore is iron.

A metal which can occur in the form of carbonate ore is zinc.

A metal which can occur in the form of halides ore is silver.

Thus, we concluded that the metallic elements which can be recovered from ores that are oxides, carbonates, halides, or sulfides are iron, zinc, silver and lead respectively.

learn more about ore:

brainly.com/question/10306443

#SPJ4

6 0
1 year ago
Convert 135 Meters to kilometers
tamaranim1 [39]

Answer:

0.135 km

Explanation:

1 km = 1000 meters

7 0
2 years ago
What net ionic equation describes the reaction when these solutions are mixed?
almond37 [142]
I think the correct answer from the choices listed above is option D. The net ionic equation that will describe the reaction of the reactants when mixed is expressed as:

<span>2PO43–(aq) + 3Ca2+(aq) → Ca3(PO4)2(s)
</span>
Hope this answers the question. Have a nice day.
7 0
3 years ago
Read 2 more answers
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