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Zarrin [17]
3 years ago
10

Name the four elements on the Periodic Table that are considered METALLOIDS?​

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:

boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At)

Explanation:

metalloids are elements that are in between the metals and nonmetals. they are in between because they have properties of both metals and nonmetals.

you can search "metalloids on the periodic table" and go to images to see what i mean :)

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Chemosynthesis and photosynthesis are both processes that produce food. <br><br> True or false?
MA_775_DIABLO [31]

Answer:

true

Explanation:

probably true

8 0
3 years ago
What is silicon used for?​
skelet666 [1.2K]

Answer:

glass

Explanation:

6 0
3 years ago
Read 2 more answers
How many grams of CO₂ can be produced from the combustion of 2.76 moles of butane according to this equation: 2 C₄H₁₀(g) + 13 O₂
Vilka [71]

Answer:

485.76 g of CO₂ can be made by this combustion

Explanation:

Combustion reaction:

2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)

If we only have the amount of butane, we assume the oxygen is the excess reagent.

Ratio is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of dioxide

Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂

We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g

5 0
3 years ago
When light strikes chlorophyll molecules, they lose electrons, which are ultimately replaced by other electrons obtained by ____
fomenos
The answer is by "splitting water".

hope this helps
5 0
3 years ago
Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
2 years ago
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