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prisoha [69]
3 years ago
8

An electron is in a region outside the nucleus. An electron

Chemistry
1 answer:
GenaCL600 [577]3 years ago
8 0

Answer:

b

Explanation:

electrons lie in the orbits of the atom. they are negatively charged and they are very small in size. they have very less mass than proton. ele tron has 1/1840 mass of proton.

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Muszę napisać wzór sumatryczny wysyłam załącznik
ioda

Cześć, nie mówię po polsku, więc zrobiłem to w Tłumaczu Google, więc przepraszam, jeśli coś brzmi śmiesznie.

chlor (VII) i tlen - ten wzór to CI_{2} O_{7}, a tlenek chloru jest bezwodnikiem kwasu nadchlorowego.

węgiel i wodór - wzór na węgiel i wodór to CnH2n + 2), jest to związek organiczny, a niższa klasyfikacja jest taka, że jest to również węglowodór aromatyczny.

Mam nadzieję, że to pomoże, błogosławionego i cudownego dnia! :-)

-Cutiepatutie

7 0
3 years ago
Why does Earthquakes occur?
Annette [7]
Shifting plates, underwater volcanoes. many different variables come in play
6 0
3 years ago
Solutions are said to be what kind of mixture? A) Pure substance B) Heterogeneous C) Homogenous D) Element
laiz [17]
Solutions are said to be C. homogeneous mixtures, composed of two or more substances. It is usually liquid, however it may be solid or gas.
7 0
3 years ago
what is the molarity of a solution prepared dissolving 317 g of CaCl2 into enough water to make 2.50 L of solution?
Minchanka [31]

Answer:

1.14 M

Explanation:

Step 1: Calculate the moles corresponding to 317 g of calcium chloride (solute)

The molar mass of calcium chloride is 110.98 g/mol.

317 g CaCl₂ × 1 mol CaCl₂/110.98 g CaCl₂ = 2.86 mol CaCl₂

Step 2: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = moles of solute / liters of solution

M = 2.86 mol / 2.50 L = 1.14 mol/L = 1.14 M

4 0
2 years ago
Will mark the brainiest later for correct answers! Please show work.
Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number 6.023\times 10^{23} of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}

a. moles in 14.08 g of C_{12}H_{22}O_{11} = \frac{14.08g}{342.3g/mol}=0.04113moles

molecules in 14.08 g of C_{12}H_{22}O_{11} = 0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}

b. moles in 17.75 g of NaCl = \frac{17.75g}{58.5g/mol}=0.3034moles

molecules in 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

formula units 17.75 g of NaCl = 0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}

c. moles in 20.06 g of  CuSO_4.5H_2O= \frac{20.06g}{249.68g/mol}=0.08034moles

formula units in 20.06 g of  CuSO_4.5H_2O= 0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}

7 0
3 years ago
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