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stich3 [128]
4 years ago
14

Consider the half reaction below.

Chemistry
1 answer:
allsm [11]4 years ago
6 0

Answer:

balancing of charges of both sides

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A sample of iron metal absorbs 4.56j of heat, upon which temperature of the sample increased from 29 degrees c to 36 degrees c.
solmaris [256]

Answer:

Mass = 1.42 g

Explanation:

Given data:

Amount of heat absorbed = 4.56 j

Initial temperature = 29°C

Final temperature = 36°C

Mass of iron metal = ?

Specific heat capacity of iron = 0.46 j/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 36°C - 29°C

ΔT = 7°C

Q = m.c. ΔT

m = Q / c. ΔT

m = 4.56 j / 0.46 j/g°C . 7°C

m = 4.56 j /3.22 j/g

m = 1.42 g

6 0
3 years ago
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant f
umka21 [38]

Answer:

1) 0.00498 mol Cu.

2) 0.00000374 mol CO₂

Explanation:

Question 1)

We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.

Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:

\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}

We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:

\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}

Cancel like terms:

=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }

Simplify:

\displaystyle  = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}

Use a calculator:

= 0.004981... \text{ mol Cu}

Since the resulting answer must have three significant figures:

= 0.00498\text{ mol Cu}

So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.

Question 2)

We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.

By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:

\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}

To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:

\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}

Cancel like terms:

\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}

Simplify:

\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}

Use a calculator:

=0.000003736...\text{ mol CO$_2$}

Since the resulting answer must have three significant figures:

= 0.00000374\text{ mol CO$_2$}

So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.

4 0
3 years ago
Read 2 more answers
How many moles of oxygen are produced when 8 moles of Al are produced?
vichka [17]
<span>how many molesof oxygen are produced when 8 mols of Al are produced? it is 6</span>
6 0
3 years ago
Read 2 more answers
¿El agua salada es una mezcla homogénea o heterogénea?
mr_godi [17]

Answer:

Homogénea

Explanation:

Una mezcla homogénea ocurre cuando se unen dos o más sustancias y no se pueden identificar después de ser unidas

En este caso no se logran identificar los compuestos

7 0
3 years ago
___au2s3 ___h2 → ___au ___h2s
liberstina [14]
The best way to balance an equation is to balance one atom at a time.
You start with two Au atoms on the left, so you know the coefficient of Au on the right has to be 2. So at first we get,
Au2S3 + H2 --> 2Au + H2S
Then, notice you have 3 sulfur atoms on the left, so you need three on the right.
Our equation becomes
Au2S3 + H2 --> 2Au + 3H2S
Lastly, we now have six hydrogen atoms on the right, and only two on the left, so we assign a three to the H2 on the left
Au2S3 + 3H2 --> 2Au + 3H2S Is the balanced final equation.

7 0
3 years ago
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