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Marysya12 [62]
3 years ago
14

What is litmus paper used for? To indicate if all the reactants have been used up to determine if a chemical reaction has occurr

ed to determine pH to detect the presence of water
Chemistry
1 answer:
rjkz [21]3 years ago
8 0
The main use of litmus is to test whether a solution is acidic or basic. Blue litmus paper turns red under acidic conditions and red litmus paper turns blue under basic or alkaline conditions, with the color change occurring over the pH range 4.5–8.3 at 25 °C (77 °F).
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When a plant _____, roots grow and a seedling begins to grow?
dusya [7]

Answer:

stem

Explanation:

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2 years ago
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What is the difference between food web and food chain and what do they have in common?
abruzzese [7]

A food web is a system of interlocking and interdependent food chains.


A food chain is a hierarchical series of organisms each dependent on the next as a source of food.


Basically, the difference of the two is that a food web contains multiple different species in the same level that eat the animal below it's status. A food chain is one after the other.

7 0
3 years ago
Please help. It's for chemistry
irinina [24]
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3 0
2 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
3 years ago
A sample of air was collected on a day when the total atmosphere pressure was
zhuklara [117]

Answer:

1. The gas law used: Dalton's law of partial pressure.

2. Pressure of nitrogen = 331 mmHg

Explanation:

From the question given above, the following data were obtained:

Total pressure (Pₜ) = 592 mmHg

Pressure of Oxygen (Pₒ) = 261 mmHg

Pressure of nitrogen (Pₙ) =?

The pressure of nitrogen in the sample can be obtained by using the Dalton's law of partial pressure. This is illustrated below:

Pₜ = Pₒ + Pₙ

592 = 261 + Pₙ

Collect like terms

592 – 261 = Pₙ

331 = Pₙ

Pₙ = 331 mmHg

Therefore, the pressure of nitrogen in the sample is 331 mmHg

5 0
3 years ago
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