385 x 42.13 x 0.079 is (consider significant figures):

Can some please answer these two questions?

alexandr402 [8]

Answer:

1. Mass of potassium (K) = 203.32 g

2. Number of mole of As = 7.53 moles

Explanation:

1. Determination of the mass of potassium (K)

Molar mass of K = 39.1 g/mol

Number of mole of K = 5.2 moles

Mass of K =.?

Mole = mass / Molar mass

5.2 = mass of K / 39.1

Cross multiply

Mass of K = 5.2 × 39.1

Mass of potassium (K) = 203.32 g

2. Determination of the number of mole of Arsenic (As)

Molar mass of As = 74.92 g/mol

Mass of As = 563.9 g

Number of mole of As =.?

Mole = mass /Molar mass

Number of mole of As = 563.9 / 74.92

Number of mole of As = 7.53 moles

8 0

3 years ago

How many moles of copper are present in 75.0g Cu? Show your complete calculation and give the answer in the correct number of si

velikii [3]

Answer:

The answer to your question is 1.2 moles of copper

Explanation:

Data

mass of copper = 75 g

moles = ?

Process

1.- Look in the periodic table for the atomic mass of copper.

Atomic mass = 63.55 g/mol

2.- Use proportions to determine the moles of copper in 75 g

63.55 g of copper -------------------- 1 mol

75 g of copper -------------------- x

Use cross multiplication

x = (75 x 1) / 63.55

x = 75 / 63.55

x = 1.18 moles ≈ 1.2 moles of copper

3 0

4 years ago

Read 2 more answers

A mass of 12.0g of carbon contains what?

tatuchka [14]

It contains one mol of carbon atoms. For the better calculating in the chemical reaction, we often use the mol as unit to describe the amount. Carbon is consists of carbon atoms. And the mol number is related to the mass of atoms. The molar mass of carbon is 12 g/mol. So the answer is 1 mol carbon atoms.

8 0

3 years ago

Determine which of the acids are Arrhenius acids, Brønsted–Lowry acids, and Lewis acids. It is possible for an acid to be of mor

tatyana61 [14]

This is an incomplete question, here is a complete question.

Determine which of the acids are Arrhenius acids, Brønsted–Lowry acids, and Lewis acids. It is possible for an acid to be of more than one type. Which acids are Arrhenius acids?

AlCl₃ (aq)

BCl₃ (aq)

HCl (aq)

Answer :

HCl is an Arrhenius acid and Bronsted Lowry acid

AlCl₃ is a Lewis-acid

BCl₃ is a Lewis-acid

Explanation :

According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions $(OH^-)$ when dissolved in water and an acid is defined as a substance which donates hydronium ions $(H_3O^+)$ in water.

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

From this we conclude that:

HCl is an Arrhenius acid because it donates hydrogen ion and also Bronsted Lowry acid because it also donate protons.

AlCl₃ is a Lewis-acid because it is a electron deficient and accept lone pair of elections.

BCl₃ is a Lewis-acid because it is a electron deficient and accept lone pair of elections.

7 0

3 years ago

If the density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cub

Tasya [4]

Answer:

1.35 × 10⁴ kg/m³ at 22 °C; 1.34 × 10⁴ kg/m³ at 100 °C

Explanation:

The cubic expansivity (γ) of a liquid is the fractional change in volume per unit change in temperature.

$\gamma = (\frac{\Delta V }{ V_0} )(\frac{1 }{ \Delta T} )$ Multiply by V₀ΔT and transpose

ΔV = γV₀ΔT

and

V = V₀ + ΔV

===============

At 0 °C

Assume you have 1 m³ of Hg

ρ = m/V Multiply by V and transpose

m = ρV

ρ = 1.36 × 10⁴ kg/m³

m = 1.36 × 10⁴ × 1 = 1.36 × 10⁴ kg

===============

At 22 °C

Assume that you have 1 m³ of Hg

γ = 180 × 10⁻⁶ K⁻¹

ΔT = 22 °C – 0 °C = 22 °C

ΔV = 180 × 10⁻⁶ × 22

ΔV = 3.96 × 10⁻³ m³ Calculate volume

V = 1 + 0.00396

V = 1.00396 m³ Calculate density

ρ = 1.36 × 10⁴/1.00396

ρ = 1.35 × 10⁴ kg/m³

===============

At 100 °C

ΔT = 100 °C – 0 °C = 100 °C

ΔV = 180 × 10⁻⁶ × 100

ΔV = 0.0180 m³ Calculate volume

V = 1 + 0.0180

V = 1.0180 m³ Calculate density

ρ = 1.36 × 10⁴/1.0180

ρ = 1.34 × 10⁴ kg/m³

4 0

4 years ago