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4 years ago

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What is true of a synthesis reaction?

1 answer:

mina [271]4 years ago

4 0

There is only one product in a synthesis reaction.

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Therefore, the independent variable was<br> and the dependent<br> variable was

den301095 [7]

Answer:

The two main variables in an experiment are the independent and dependent variable. An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.

5 0

3 years ago

Read 2 more answers

Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.

Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

$C(gr) +H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The combustion of liquid acetic acid is given as,

$CH_{3} COOH(l) + 2O(g)$ → $2CO_{2}((g) +2H_{2} O(l)$ ΔH =-873

learn more about balancing chemical equation

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5 0

2 years ago

Please help me with the questions?

weqwewe [10]

Answer:

1. Forecast

2. Cold front

3. Shelf Cloud

4. Barometer

5. Pressure

6 Prevailing Westerlies

7. Thermometer

8. Weather

6 0

3 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

A neutral particle formed when atoms share electrons.

Yakvenalex [24]

Answer:

A

ion

ionic bond

compound

convalent bond

electrons

molecule

Explanation:

6 0

3 years ago