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3 years ago

In order for convection to transfer heat, particles need to

Chemistry

2 answers:

inna [77]3 years ago

7 0

It’s b make contact with the heat source

scZoUnD [109]3 years ago

7 0

The answer is B, for something to become hot like a pan it needs to be touching something hot

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The sculpted pinnacles in Bryce Canyon National Park are the result of _____. A. differential weathering B. frost wedging C. exf

KonstantinChe [14]

Answer:

B. frost wedging Hope i helped!

Explanation:

5 0

3 years ago

Read 2 more answers

A solution was prepared by dissolving 0.05000 mol of sucrose in 100.0 kg of water. Find

Licemer1 [7]

Answer:

0.0005mol/kg

Explanation:

Molality = mole of solute/mass of solvent (kg)

From the question, we obtained the following data:

Mole of solute = 0.05mole

Mass of solvent = 100kg

Molality = 0.05/100

Molality = 0.0005mol/kg

The molality of the solution is 0.0005mol/kg

8 0

4 years ago

Read 2 more answers

How many grams of CO2 are produced from 6.7 L of O2 gas at STP?

Tasya [4]

<h3>Answer:</h3>

13 g CO₂

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

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

Using Dimensional Analysis

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] 6.7 L O₂

[Solve] g O₂

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

[STP] 22.4 L = 1 mol

[PT] Molar Mass of O: 16.00 g/mol

[PT] Molar Mass of C: 12.01 g/mol

Molar Mass of CO₂: 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 6.7 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2})(\frac{44.01 \ g \ O_2}{1 \ mol \ O_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 13.1637 \ g \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

13.1637 g CO₂ ≈ 13 g CO₂

5 0

3 years ago

HelpUsing a 2-cm-thick piece of cardboard over a radiation source would be mosteffective for protecting against which type of ra

Lisa [10]

It would protect best against C) Alpha radiation, as Beta radiation is stopped by lighter metals such as aluminium, and Gamma radiation can only be stopped by heavier metals such as lead.

3 0

4 years ago

Read 2 more answers

A sample of an unknown gas effuses in 14.4 min. An equal volume of H2 in the same apparatus under the same conditions effuses in

Elena-2011 [213]

Answer:- molar mass of the unknown gas is 71.5 gram per mol.

Solution:- From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

When we compare the effusion rates of two gases then the formula for Graham's law is:

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

In this formula, V stands for volume and M stands for molar mass

Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$

let's say in formula, subscript 1 is for hydrogen gas and 2 is for the unknown gas.

Molar mass of hydrogen is 2.02 grams per mol and the time taken to effuse it is 2.42 min. The time taken to effuse the unknown gas is 14.4 min and we are asked to calculate it's molar mass. let's plug in the values in the formula:

$\frac{14.4}{2.42}=\sqrt{\frac{M_2}{2.02}}$

$5.95=\sqrt{\frac{M_2}{2.02}}$

doing squares to both sides:

$35.4=\frac{M_2}{2.02}$

$M_2=35.4*2.02$

$M_2=71.5$

So, the molar mass of the unknown gas is 71.5 grams per mol.

4 0

3 years ago