I’ll give 5 stars and Brainliest plus this is 20 points lol

Which of the following types of radiation can be shielded by piece of paper or fabric

Keith_Richards [23]

There are 3 major types of radiation. The Alpha, Beta and Gamma radiation. It is the Alpha radiation that can be stopped by a piece of paper or fabric. Beta on the other hand can be stopped by plastic or a thin sheet of metal and Gamma by a piece of aluminum foil or lead.

8 0

3 years ago

What items are true about a block of ice at -10°C as you continue to apply heat

zhenek [66]

1. its temperature will rise continuously until it melts

I don't believe that any of the other answers are correct because it can not stay at a certain temperature if it is melting

5 0

3 years ago

How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?

faltersainse [42]

Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

Volume of the solution (V): 500 mL

Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

4 0

3 years ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

<u>Step 2: Identify Conversions</u>

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

3 years ago

Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.

Elanso [62]

Answer:

Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.

Explanation:

1 mole of any compound contains 6.023 × 10²³ molecules.

molecular weight of NaCl is 23 + 35.5 = 58.5 g.

so, 58.5 grams of NaCl makes 1 mole

⇒ 14.5 g of NaCl = $\frac{14.5}{58.5}$ = 0.248 moles.

⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules

= 1.49 × 10²³ molecules.

And 1 molecule contains 1 Na ion and 1 Cl ion.

⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.

6 0

4 years ago