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3 years ago

1. This is very important in preventing the spread of pathogens in the kitchen.

1 answer:

maks197457 [2]3 years ago

6 0

Sanitising is very important in a house hold but most importantly in the kitchen we have to always keep things clean and sanitise for the sake of our health and the people around us

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In addition to not causing damage to the sample, what is another advantage of using a microspectrophotometer to analyze fibers?

Tems11 [23]

Another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.

<h3>What is a microspectrophotometer?</h3>

Microspectrophotometry is a biological technique used to measure the absorption or transmission spectrum of a solid or liquid material in either transmitted or reflected light.

Microspectrophotometry can also measure the emission of light by a sample, which is usually small as the micro implies.

One advantage of microspectrophotometry is that the sample does not get damaged. However,

However, another advantage of advantage of using a microspectrophotometer to analyze fibers asides not causing damage to the sample is that the sample can be quite small.

Learn more about microspectrophotometry at: brainly.com/question/5832827

5 0

2 years ago

How is oxygen and nitrogen alike

11111nata11111 [884]

Answer:

They are both colorless, odorless, and tasteless. They have the same number of valence electrons too. And unbalanced electrons in their valence shell.

Explanation:

3 0

3 years ago

Read 2 more answers

What happens to gas particles when they are compressed?

Anuta_ua [19.1K]

Answer:

3)Some gas molecules move further apart and some move closer together.

Explanation:

because When more gas particles enter a container, there is less space for the particles to spread out, and they become compressed. The particles exert more force on the interior volume of the container. This force is called pressure. There are several units used to express pressure.

4 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

slega [8]

<h2>Answer:</h2> $1.33*10^{-2}grams$

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

$\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}$

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

$\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}$

Determine the mass of water produced

$\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}$

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

5 0

1 year ago