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3 years ago

I need to know how do awnser this pease help.

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

6 0

If you imagined the line BCD as a line and ACE as a transversal, this would make angle ACB and DCE alternate interior angles. Since all three angles in a triangle, add up to 180 degrees, this means that the side lengths should be similar or congruent to accommodate the angle measure.

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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 between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

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 : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

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<span>The answer is curie.

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What is the percent error of a length measurement of .454cm if the correct value is 0.546cm?

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17% is the percent error

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2 years ago

A 250.0-ml sample of ammonia, nh3 (g), exerts a pressure of 833 torr at 42.4 °c. what mass of ammonia is in the container

Lelu [443]

To solve this, let's assume ideal gas behavior.

PV=nRT
Let's solve for n. Convert units to SI units first.

Pressure = 833 torr(101325 Pa/760 torr) = 111,057.53 Pa
Volume = 250 mL(1 L/1000 mL)(1 m³/1000 L) = 2.5×10⁻⁴ m³
Temperature = 42.4 + 273 = 315.4 K

n = (8,314 J/mol·K)(315.4 K)/(111057.53 Pa)(2.5×10⁻⁴ m³)
n = 94.45 mol

The molar mass of ammonia is 17.031 g/mol.
Mass = 94.45*17.031 = <em>1,608.51 g ammonia</em>

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3 years ago