All categories

3 years ago

If an element has an atomic number of 35 and an atomic mass of 84, how many protons does this element have?

Chemistry

1 answer:

vfiekz [6]3 years ago

7 0

Answer:

Explanation:

because the atomic number is the same as the protons number

You might be interested in

Dominic and Eva are using the same type of stopwatches to measure the time it takes for a chemical reaction to occur. They each

SIZIF [17.4K]

They did not repeat their tests multiple times.

Reason being is using the same stopwatch doesn’t matter and changing the reactants won’t help, and you without a doubt need more than one person collecting the data so,

They did not repeat their tests multiple times is the answer.

4 0

2 years ago

Why is the base of an energy pyramid wider than the top

Zepler [3.9K]

It is due to a lack of providable energy to the next trophic level of the energy pyramid. Primary consumers only obtain around 10% of energy that producers have, and the energy depletes as you move further and further up the food chain, until you reach the tertiary consumers which have the least amount of energy at their disposal, meaning there are less of them than other organisms further down the energy pyramid.

Hope this helped!

5 0

3 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

3 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

3.77 mol K₃N

Step 2: Identify Conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

Step 3: Convert

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

495.039 g K₃N ≈ 495 g K₃N

5 0

3 years ago

What is a glass of iced tea containing excess sugar at the bottom said to be?

Pavel [41]

The answer is saturated

5 0

3 years ago