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andreev551 [17]

3 years ago

7

Which explains why radioactive waste is so difficult to get rid of? A) Low electrical charge Eliminate B) High electrical charge

C) Long half-life of waste products D) Short half-life of waste products

2 answers:

stepladder [879]3 years ago

4 0

I believe the answer is C. long half-life of waste products

matrenka [14]3 years ago

4 0

Answer:

C) Long half-life of waste products

Explanation:

I got it right

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1) A light bulb takes in 30 of energy per second. It transfers 3j as use

natta225 [31]

Answer:

$\boxed{\text{10 \%}}$

Explanation:

The formula for efficiency is

$\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}$

Data:

Useful energy = 3 J

Energy input = 30 J

Calculation:

$\begin{array}{rcl}\eta & = & \dfrac{\text{3 J}}{\text{30 J}} \times 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}$

8 0

3 years ago

If the temperature is below zero, what will the precipitation most likely be?

Leviafan [203]

Answer: C Snow

Explanation:

Because the temperature is low and it is below freezing temperature. Sorry if I am wrong.

7 0

2 years ago

How much power is used if a force of 54 newtons is used to pull a wagon a distance of 10 meters in 6 seconds? 3,240 W 90 W 32.4

timama [110]

Power is defined in a mathematical expression as P = F x v where F is in N and v is in m/s. From the given equation, the v = d/t which is v = 10/ 6, then substituting the answer to the power formula W = 54 N (10/6 m/s) = 90 Watts.

3 0

3 years ago

Read 2 more answers

An irregularly-shaped piece of copper (Cu) has a mass of 55.0 grams. What is the volume in cm³ of this piece of copper if its de

vredina [299]

Answer:

<h2>6.14 cm³</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question we have

$volume = \frac{55}{8.96} \\ = 6.138392...$

We have the final answer as

<h3>6.14 cm³</h3>

Hope this helps you

8 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

2 years ago