Which element has exactly ﬁve electrons in the highest principal energy level (the outer shell)?

What is used up in and stops a chemical reaction?

frutty [35]

Answer:What is used up in and stops a chemical reaction? (a) percent yield (b) limiting reactant (c) theoretical yield (d) excess reactant

The correct option is (b)

Explanation:

A chemical reaction is a chemical combination,rearrangement or disintegration of chemical substance(s) (called reactants) to form new chemical substance(s) (called products).

A limiting reactant is the reactant that is totally used up during the course of a reaction. When this reactant is used up, it limits the amount of products formed and eventually stops the chemical reaction.

Explanation:

The limiting reagent is the reactant that is used up completely. This stops the reaction and no further products are made. ... This method is most useful when there are only two reactants. The limiting reagent can also be derived by comparing the amount of products that can be formed from each reactant.

hope this helps have a nice day❤️

6 0

3 years ago

Ehat coefficient would go in the box

Natalija [7]

Answer:

1.nhl

Explanation:

8 0

2 years ago

Can some one work out this equation for me :)

Law Incorporation [45]

Yo sup??

CaCO3 + 2 HCl ---------> CaCl2 + H2O + CO2

This is the balanced chemical equation

Hope this helps

4 0

2 years ago

What are the two types of numbers in experimental calculations?

FinnZ [79.3K]

The logistics of a proposed larger study

Gain familiarity with the experimental material,

Ensure that treatments are not obviously excessively mild or severe

Check that staff are sufficiently well trained in the necessary procedures

Ensure that all steps in a proposed future experiment are feasible.

Gain some information on variability, although this will not usually be sufficiently reliable to form the basis of power analysis calculations of sample size.

Exploratory experiments can be used to generate data with which to develop hypotheses for future testing. They may “work” or “not work”. They may have no clearly stated hypothesis (“let’s see what happens if..” is not a valid hypothesis on which to base an experiment).

Often they will measure many outcomes (characters). Picking out “interesting looking differences” (known as data snooping) and then doing a hypothesis test to see if the differences are statistically significant will lead to serious overestimation of the magnitude of a response and excessive numbers of false positive results. Such differences should always be tested in a controlled experiment where the hypothesis is stated a priori before the results are published.

Depending on the nature of the data, statistical analysis will often be done using an analysis of variance (ANOVA)

Confirmatory experiments are used to test some relatively simple hypothesis stated a priori. This is the type of experiment mainly considered in this web site.

The basic principles are:

Experiments involve comparisons between two or more groups

Their aim is to test a “null hypothesis” that there is no difference among the groups for the specified outcome.

If the null hypothesis is rejected at a certain level of probability (often 5%) this means that the probability of getting a result as extreme as this or more extreme in the absence of a true effect is 5% (assuming also that the experiment has been properly conducted). So it is assumed that such a difference is likely to be the result of the treatment. But, it could be a false positive resulting from sampling variation.

Failure to reject the null hypothesis does not mean that the treatment has no effect, only that if there is a real effect this experiment failed to detect it. “Absence of evidence is not evidence of absence”.

Experimental subjects need to be independently replicated because individuals (of whatever type) vary. Two subjects can normally be regarded as being independent if they can theoretically receive different treatments.

Subjects need to be assigned to groups, held in the animal house and measured at random in order to minimise the chance of bias (a systematic difference between groups)

As far as possible the experimenter should be “blind” with respect to the treatment group in order to minimise bias.

The experiments need to be powerful, i.e. they should have a high probability of detecting an effect of clinical or scientific importance if it is present.

In many cases a formal experimental designsuch as a “completely randomised”, “randomised block”, “Latin square” etc. design will be used.

In most cases it is useful if the experiment has a wide range of applicability. In other words the results should hold true under a range of different conditions (different strains, both sexes, different diets, different environments etc.). At least some of these factors should be explored using factorial and randomised block designs.

Experiments to explore relationships between variables. A typical example would be a growth curve or a dose-response relationship. In these experiments the aim is often to test whether the two variables are associated, and if so, what is the nature of that relationship. The typical statistical analysis involves correlation and/or regression.

8 0

3 years ago

In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The

Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

Step 1: Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

Step 2: Calculate mass of NaNO2

(40/100)*1.05 = 0.42 grams

Step 3: Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

Step 4: Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

Step 5: Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

Step 6: Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 = 0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

4 0

3 years ago