Give the ion notation for an atom with 34 protons and 36 electrons

Which of the following best describes an electron?

GenaCL600 [577]

B. It has a negative charge and much less mass than a proton.

3 0

2 years ago

Read 2 more answers

Someone answer the limited regent...

OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

Using Dimensional Analysis

Chemistry - Atomic Structure

Reading a Periodic Table

Explanation:

Step 1: Define

[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given] 10 g Cu

Step 2: Identify Conversions

[RxN] 1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

Step 3: Stoichiometry

$10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} )$ = 16.974 g Ag

Step 4: Check

We are given 1 sig fig. Follow sig fig rules and round.

16.974 g Ag ≈ 20 g Ag

6 0

2 years ago

What is the percent error when a student measures the volume to be 19.3 liters when

katrin2010 [14]

Answer:

The answer is

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual volume = 22.4 L

error = 22.4 - 19.3 = 3.1

The percentage error is

$P(\%) = \frac{3.1}{22.4} \times 100 \\ = 13.839285714...$

We have the final answer as

Hope this helps you

7 0

2 years ago

At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

6 0

2 years ago

Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.089 m in

Anuta_ua [19.1K]

Answer: 14

Explanation:

6 0

2 years ago