An isotope contains 25 protons, 30 neutrons, and 23

Which of the following would be considered a compound?<br> He<br> CI<br> NaCl<br> B

kobusy [5.1K]

Answer:

NaCl

Explanation:

because He which is helium gas its an element alone,

Cl which is chlorine is a single element,

B which is Boron and its a single element,

but NaCl which is simply called table sugar is the combination between Na, sodium and Cl, chlorine , so its called a compound not an element.

5 0

2 years ago

The amount of. moisture in the air is highest when relative humanity is

gogolik [260]

Answer:

Saturation is the maximum amount of water vapor in the air at an existing temperature and pressure. Air is said to be saturated at 100 percent relative humidity when it contains the maximum amount of moisture possible at that specific temperature. Dew point is the temperature when air reaches 100% relative humidity.

6 0

2 years ago

List the number of

rjkz [21]

Answer:

beryllium: 2 valence electrons

nitrogen: 5 valence electrons

oxygen: 6 valence electrons

fluorine: 7 valence electrons

magnesium: 2+ valence electrons

phosphorus: 5 valence electrons

sulfur: 6 valence electrons

chlorine: 7 valence electrons

Explanation:

6 0

2 years ago

What is a primary consumer

Ket [755]

Answer:

They eat producers

Explanation:

for example, any herbivore

3 0

3 years ago

Read 2 more answers

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago