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3 years ago

I really need help with this, someone help me please.

Chemistry

1 answer:

Pepsi [2]3 years ago

6 0

Its C. It has to be equivalent to the number divided or multiplied by 5.

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The electron configuration belonging to the atom with the highest second ionization energy is ________.

Gemiola [76]

Answer:

Element Lithium

Explanation:

The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals

It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.

Its atomic number is 3. The electronic configuration is 1s2 2S1

5 0

3 years ago

Year/ Distance (cm)

luda_lava [24]

Answer:

the answer is C

Explanation:

6.7 to 13.2 then look at the numbers they go up but not a lot each time

5 0

3 years ago

A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the

yarga [219]

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

5 0

2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

Please help <br> Show your work

Yuliya22 [10]

Answer:

Explanation: 2+1=21

6 0

2 years ago