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Klio2033 [76]
3 years ago
6

Is smelting copper is a chemical change or physical change?

Chemistry
1 answer:
Grace [21]3 years ago
8 0
Smelting copper is a chemical change
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C6H12O6 + 6O2 → 6CO2 + 6H2O is the chemical equation for cellular respiration. What are the products in this reaction?
maks197457 [2]

Answer:

The products are: A) CO2, H2O

Explanation:

Those products that are seen on the right side of the reaction (that is, those substances that are generated from the reagents). In this case they are carbon dioxide and water.

The general equation of cellular respiration is:

C6H1206 + 602 -> 36 ATP + 6CO2 + 6H20

4 0
3 years ago
Margret claimed that she needed 5.45 moles of Potassium Chloride, KCl to complete her experiment. How many grams of KCl does she
zubka84 [21]

Answer:

loiuygfcx

Explanation:

3 0
2 years ago
If a solution containing
tankabanditka [31]

Answer:

Hg(NO₃)₂(aq) + Na₂SO₄(aq)   →   2NaNO₃(aq) + HgSO₄(s)

Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

                                        = 0.1172 x 296.65

                                        = 34.757g

Explanation:

6 0
3 years ago
What is scientific notation?
lora16 [44]

Answer:

C.

Explanation:

Scientific notation is a way of writing small and large numbers. It is a way of expressing a standard form into a scientific form.

8 0
3 years ago
Read 2 more answers
In the summer of 2016, the city of Columbus Dublin Road Water Treatment plan exceeded the regulatory level of nitrate, which is
Verizon [17]

Answer:

a) 10 mg NO_3^-/L

b) 1.61*10^{-4}mol NO_3^-/L

c) 2.26 mg N/L

Carbon: C=26.64 \frac{mg C}{L}

Explanation:

<u>Nitrate</u>

First of all, is important to know that:

1 ppm=1 mg/L

a) 10 ppm of nitrate (NO_3^-) is equal to 10 mg NO_3^-/L

b) The molecular weight of nitrate is 62 g NO_3^-/mol

10 mg NO_3^-/L=0.01 g NO_3^-/L

\frac{0.01 g NO_3^-/L}{62 g NO_3^-/mol}=1.61*10^{-4} mol NO_3^-/L

c) Nitrate has 14 mg of N per 62 mg of NO3

10 mg NO_3^-/L*\frac{14 mg N}{62 mg NO_3^-}=2.26 mg N/L

<u>Carbon</u>

Carbonate has 12 mg of C per 60 mg of CO_3^{-2}

Bicarbonate has 12 mg of C per 61 mg of HCO_3^{-}

C=24 \frac{mg CO_3^{-2}}{L}*\frac{12 mg C}{60 mg CO_3^{-2}}+111 \frac{mg HCO_3^{-}}{L}*\frac{12 mg C}{61 mg HCO_3^{-}}

C=26.64 \frac{mg C}{L}

8 0
2 years ago
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