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KiRa [710]
3 years ago
15

In the image below, the truck and bus have the same mass and are

Chemistry
1 answer:
REY [17]3 years ago
5 0

Answer:

B)They have the same kinetic energy but different gravitational potential energies.

Explanation:

I don't know the exact explanation but I just took the test.

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agasfer [191]

Answer:

option c hope it will help u

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2 years ago
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Which of the following experiments would best test viscosity?
Juli2301 [7.4K]
C. pouring honey on a plate so the density of thickness and stickyness would 
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3 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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Look at the image. What type of eclipse is being shown in this image?
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3 years ago
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What is the mass in grams of BaCl2 that is needed to prepare 200 mL of a 0.500 M solution
DENIUS [597]

Answer:

= 20.82 g of BaCl2

Explanation:

Given,

Volume = 200 mL

Molarity = 0.500 M

Therefore;

Moles = molarity × volume

          = 0.2 L × 0.5 M

          = 0.1 mole

But; molar mass of BaCl2 is 208.236 g/mole

Therefore; 0.1 mole of BaCl2 will be equivalent to;

  = 208.236 g/mol x 0.1 mol

  = 20.82 g

Therefore, the mass of BaCl2 in grams required is 20.82 g

6 0
3 years ago
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