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aliina [53]
2 years ago
10

What combines to form compounds? A. atoms of elements B. molecules C. minerals D. valences

Chemistry
2 answers:
hammer [34]2 years ago
8 0
D would be correct (i did this Assessment today)<span />
wlad13 [49]2 years ago
5 0

the correct answer would be

<u><em>A. atoms of elements </em></u>

i took the test

<em>please mark as brainliest </em>

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The answer is E it’s killing all oxygen that is around you
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The movement of water from the atmosphere to the ground is called which of the following?
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What is the empirical formula for Hg2(NO3)2
den301095 [7]
<span>Answer: HgNO₃
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<span>Explanation:
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The empirical formula is the formula that shows the ratio of the atoms in its simplest form, this is using the smallest whole numbers.
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<span>The empirical formula may or may not be the same molecular formula.
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<span>In this case you are given the molecular formula Hg₂(NO₃)₂. Since, the ratio of the atoms of Hg, N, and O is 2: 2: 6, respectively, the same ratio is expressed if you divide by the greatest common factor (GCF).
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8 0
3 years ago
Write the chemical formula of the following:
Elan Coil [88]

Answer:

1.Ca(OH)2

2.Mg(OH)2

3.Al2SO4

4.Na2CO3

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5 0
3 years ago
A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

7 0
2 years ago
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