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2 years ago

Exam number: 007170RR

Chemistry

1 answer:

AleksandrR [38]2 years ago

5 0

Answer:

A open

Explanation:

I am not sure but i think it is!

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Tems11 [23]

Answer:

year 1 is 5.5%

year 2 is 7.5%

year 3 is 10.2%

Explanation:

since,

length of the transect covered in seaweed / total lenth of transect x 100

then,

0.55 / 10.0 x 100 = 5.5

and

0.75 / 10.0 x 100 = 7.5

and

1.02 / 10.0 x 100 = 10.2

you could also just move the decimal to the right once

4 0

2 years ago

How many grams of FeCl 3 are in 250. mL of a 0.100 M solution?

Naddika [18.5K]

Answer:

The correct answer is option B

Explanation:

$$Molarity=\frac{Weight}{Molecular \,weight} \frac{1000}{V(in \, ml)}$

Given values,

Molarity of $FeCl_3=0.100M$

Volume of solution, $V=250ml$

Molecular weight of $FeCl_3=162.2$

Substituting this values in Molarity formula, we get

$0.1=\frac{weight}{162.2} \times\frac{1000}{250} $\\$\Rightarrow 16.22=weight\times4$\\$\Rightarrow weight=\frac{16.22}{4} $\\$\therefore weight=4.06g$

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3 years ago

Which of these choices is required by sustainable agriculture but not by<br> organic agriculture?

zvonat [6]

Answer:

it is (a) USDA certification

Explanation:

apex

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2 years ago

What does the thumping on the chest signal mean?

ankoles [38]

It means that you are the brother or friend of the person doing the thumping

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2 years ago

Read 2 more answers

g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint

diamong [38]

Answer:

The value is $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

The molar mass of $CaCO_3$ is $MW = 100.085 \ g/mol$

The total mass is $m_g = 12.3 \ g$

The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

The percentage of calcium in calcite is mathematically represented as

$C = \frac{40.07}{100.085} * 100$

$C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

$m= \frac{m_g}{5}$

$m= \frac{12.3}{5}$

$m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

$m_c = 2.46 * \frac{40.04}{100}$

$m_c = 0.985 \ g$

The uncertainty of mass of a single sample is mathematically represented as

$k = \frac{\Delta g }{5}$

$k = \frac{0.1 }{5}$

$k = 0.02\ g$

The uncertainty of mass of calcium in a single sample is mathematically represent

$G = \frac{0.02 * 40.04}{ 100}$

$G = 0.00801 \ g$

Generally the average mass of calcium in each sample is

$L = 0.985 \pm 0.00801$

6 0

2 years ago