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4 years ago

How would I go about solving this problem?

1 answer:

4 0

The first thing you need to do with problem of this kind is to write the chemical equation. This would help you in calculations later. First, convert how many moles are there in 1.75g of carbon dioxide. Then, from the chemical equation you will know how many moles of sodium bicarbonate formed with 1 mole of carbon dioxide. Then convert the moles of sodium bicarbonate to grams using the molecular weight of the compound. You now have your answer.

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Which of the following values displays the pH of an acid? A. 3 B. 12

tester [92]

A pH of 7 or lower makes something an acid so it would be 3

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3 years ago

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What is an indication that a lake is healthy?

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Answer:

Explanation:

As bioindicators are the organism that indicate or monitor the health of the environment

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3 years ago

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Name of th molecule 1. CH3CH2CHClCHBrCH3 2.C=C-CH3 CH3CH=CHCH2

Grace [21]

Answer:

1: 2-bromo-3-chloropentane

Explanation:

find longest carbon chain =5

place the Br and Cl on the carbon chain

follow naming rules I guess

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3 years ago

What is the molarity of a solution that contains 4.87 g KCI (74.55 g/mol) dissolved in in 500.0 ml water?

lbvjy [14]

Answer:

Molarity = 0.13M

Explanation:

n=m/M = 4.87/74.55= 0.065mol, V= 500ml= 0.5dm3

Applying

n= C×V

0.065= C×0.5

C= 0.13M

8 0

3 years ago

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago