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Ivenika [448]
3 years ago
14

Is it okay ?off it's difficult to find a pic for a boy ​

Chemistry
2 answers:
nalin [4]3 years ago
7 0

Answer:

Mara questions ma send Karo jaldi. or bi

Andrew [12]3 years ago
7 0

Answer:

it's not difficult

search on pin

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3 In the section describing protozoa, the author states: "Protozoa are germs that live in almost all soil, including the hot des
Gnesinka [82]

Answer:

I believe the answer is A.

Explanation:

This section states that protozoa live in all bodies of water therefore B can not be an answer because it says protozoa dose not live in any body of water.

I hope this answers your question.

7 0
3 years ago
What is test for ions?​
fomenos
It is possible to use a flame test to detect the presence of an alkali metal ion. A cleaned, moistened flame test wire is dipped into a solid sample of the compound. It is then put into the edge of a blue Bunsen flame. The flame colour produced indicates which alkali metal ion is present in the compound.






Hope it works plz mark me as brainliest
7 0
3 years ago
Read 2 more answers
Which part of Earth most likely has the highest albedo value?
andre [41]
Uhhh pretty sure B


Happy New Years
3 0
3 years ago
If element X has 49 protons, how many electrons does it have? Your Answer:
finlep [7]

It has 49 electrons as well. The amount of protons and electrons are equal.

8 0
4 years ago
If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

8 0
3 years ago
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