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3 years ago

In the chemical formula, is the metal or nonmetal symbol written first?

1 answer:

6 0

<span>Chemical formulas are written from the most metllic
for example group Alkali Metals are the most metallic, so they will be written first.</span>

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Will mark as Brainliest.

xeze [42]

Spontaneous reaction.

3 0

3 years ago

Calculate the molarity of the sodium acetate solution as described below.

Airida [17]

Answer:

This question is incomplete.

Explanation:

This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;

Molarity = number of moles (n) / volume (in liter or dm³)

To calculate the number of moles of NaC₂H₃O₂, we say

number of moles (n) =

given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂

The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000

The unit for Molarity is M (Molar concentration), mol/L or mol/dm³

4 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

When 12 moles of o2 react with 1.1 mole of c10h8 what is the limiting reactant?

Trava [24]

1.1 moles of C10H8 is the limiting reagent in the reaction between reaction C10H8 and O2
.
C10H8 + 12O2 ----> 10CO2 + 4H2O

C10H8 is the limiting reagent since 1.1 moles of C10H8 is totally consumed during the reaction

6 0

3 years ago

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Calculate the approximate volume of a 0.500 mol sample of gas at 15°C and a pressure of 1.10 atm.

Marianna [84]

T=15+273=298k
P=1.1atm
n=0.5mol

Apply ideal gas equation

PV=nRT
V=nRT/P
V=0.5(8.314)(298)/1.1
V=1126.16mL
V=1.12L

3 0

2 years ago

Read 2 more answers