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madam [21]
3 years ago
5

Identify the control group the experimental group the independent variable and the dependent variable

Chemistry
1 answer:
denis-greek [22]3 years ago
3 0
Control group: 50 dogs continuing their normal diet
Experiments group: 50 dogs chosen to eat the new food
Independent variable: dog food
Dependent variable: the dogs’ weight
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VMariaS [17]

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acid or sunlight

Explanation:

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In determining nuclear binding energy, what is Einstein's equation used to do?
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What is 20 times 200000000
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2 years ago
Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-
hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written betweenZn^{2+} (Positive ion)and F^{-} (Negative ion)

First Formula would be ZnF_{2}

Second empirical formula is between Zn^{2+}(Positive ion) and O^{2-}(Negative ion)

Second Formula would be Zn_{2}O_{2}

Note : When the subscript are same they get cancel out, so Zn_{2}O_{2} would be written as ZnO

Third empirical formula is between Ni^{4+}(Positive ion) and F^{-}(Negative ion)

Third Formula would be :NiF_{4}

Forth empirical formula is between Ni^{4+}(Positive ion)and O^{2-}(negative ion)

Forth Formula would be : Ni_{2}O_{4} or NiO_{2}

Note- The subscript will be simplified and the formula will be written as NiO_{2}.

The empirical formula of four binary ionic compounds are : ZnF_{2}, ZnO, NiF_{4},NiO_{2}


8 0
3 years ago
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A 1.3 g sample of a substance is heated from 0°C to 45°C and is found to have absorbed 45 j of heat. What is the specific heat o
wlad13 [49]

Q=mcat

45=(1.3)(c)(45)

45=58.5c

.769=c

~.77

The answer is .77

5 0
3 years ago
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