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frez [133]
2 years ago
15

Not sure what to do can you help

Chemistry
1 answer:
NeTakaya2 years ago
7 0

sorry i don't know the answer i'm really sorry

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The boiling point of a substance is tested. After 10 tests, the result is given as 37+/−3°C. Which conclusion can be drawn from
VMariaS [17]

Answer:

Explanation:

The actual boiling point is probably between 34C and 40C.

6 0
1 year ago
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The SI unit of pressure is the ______________.
steposvetlana [31]

The answer is Pascal

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3 years ago
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Copper chloride bihydrate. What might this name tell you about the atoms in this compound
lozanna [386]

Answer : The atoms in this compound are Copper(Cu), Chlorine(Cl), Hydrogen(H), Oxygen(O).

Explanation :

The given compound is copper chloride bi-hydrate which is also called as copper (II) chloride dihydrate as it contains two water of crystallisation.

The formula of copper chloride bi-hydrate is CuCl_2.2H_2O.

Therefore, there are 4 atoms in this compound and they are Copper(Cu), Chlorine(Cl), Hydrogen(H) and Oxygen(O).

6 0
2 years ago
if one mole of calcium carbonate (the limiting reactant) is used, how much calcium chloride should the reaction produce? look at
salantis [7]

Answer:

The answer to your question is <u>111 g of CaCl₂</u>

Explanation:

Reaction

                 2HCl  +  CaCO₃   ⇒    CaCl₂  +  CO₂  +  H₂O

Process

1.- Calculate the molecular mass of Calcium carbonate and calcium chloride

CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g

CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g

2.- Calculate the amount of calcium chloride produced using proportions.

The proportion CaCO₃ to CaCl₂ is   1 : 1.

                      100 g of CaCO₃  ------------- 111 g of CaCl₂

Then 111g of CaCl₂ will be produced.

5 0
3 years ago
If you have 3 moles of CH4 , how many moles of CO2 are produced?
k0ka [10]

Answer:

Since you are producing 3.6 mol CO2, you can calculate the starting moles of CH4 with the simple mole-to-mole ratio: 1 mol CH4 / 1 mol CO2 as a conversion factor. Taking 3.6 mol CO2 x 1 mol CH4 / 1 mol CO2 = 3.6 mol CH4 (after canceling out the moles of CO2 on the top and bottom of the calculation)

Explanation:

7 0
2 years ago
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