They then used RStudio to construct a linear model to predict the acetic acid concentration in the red wine. Below is the summary output for this model.summary(wine.lm)## ## Call:## lm(formula = acetic.acid ~ citric.acid)## ## Residuals:## Min 1Q Median 3Q Max ## -0.16519 -0.06360 -0.02120 0.04675 0.29145 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 0.49989 0.01800 27.77 <2e-16 ***## citric.acid -0.47058 0.04195 -11.22 <2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 0.09254 on 104 degrees of freedom## Multiple R-squared: 0.5475, Adjusted R-squared: 0.5431 ## F-statistic: 125.8 on 1 and 104 DF, p-value: < 2.2e-16Use the drop-down menus and numerical answer boxes to complete the equation of this linear model. Round numbers to 3dp.Answer == Answer ++ Answer ×× AnswerQuestion 2.3 (1 mark)based on this linear model, which ONE of the following statements is correct?A red wine that has no citric acid is expected to have 0.5 g/dm33 of acetic acidA red wine that has 1 g/dm33 more citric acid than another is expected to have 0.5 g/dm33 more acetic acidA red wine that has no acetic acid is expected to have 0.5 g/dm33 of citric acidA red wine that has 1 g/dm33 more acetic acid than another is expected to have 0.471 g/dm33 less citric acidQuestion 2.4 (2 marks)One of the bottles of red wines had 0.08 g/dm33 of citric acid and 0.62 g/dm33 of tartaric acid.The value of the residual for this bottle of red wine is: Answer (round your answer to 3dp)

A team of researchers studying red wine were interested in using a linear model to predict the concentration of acetic acid based on the concentration of citric acid found in the wine.They randomly selected 106 bottles of red wine. Concentrations are measured in g/dm33. The amount of citric acid found in these wines ranged from 0.05 g/dm33 up to 1 g/dm33.Question 2.1 (1 mark)A correlation coefficient of -0.74 was calculated between the acetic and citric acids concentrations. Which ONE of the following plots is best described by this correlation coefficient?Plot DPlot BPlot CPlot AQuestion 2.2 (4 marks)They then used RStudio to construct a linear model to predict the acetic acid concentration in the red wine. Below is the summary output for this model.summary(wine.lm)## ## Call:## lm(formula = acetic.acid ~ citric.acid)## ## Residuals:## Min 1Q Median 3Q Max ## -0.16519 -0.06360 -0.02120 0.04675 0.29145 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 0.49989 0.01800 27.77 <2e-16 ***## citric.acid -0.47058 0.04195 -11.22 <2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 0.09254 on 104 degrees of freedom## Multiple R-squared: 0.5475, Adjusted R-squared: 0.5431 ## F-statistic: 125.8 on 1 and 104 DF, p-value: < 2.2e-16Use the drop-down menus and numerical answer boxes to complete the equation of this linear model. Round numbers to 3dp.Answer == Answer ++ Answer ×× AnswerQuestion 2.3 (1 mark)based on this linear model, which ONE of the following statements is correct?A red wine that has no citric acid is expected to have 0.5 g/dm33 of acetic acidA red wine that has 1 g/dm33 more citric acid than another is expected to have 0.5 g/dm33 more acetic acidA red wine that has no acetic acid is expected to have 0.5 g/dm33 of citric acidA red wine that has 1 g/dm33 more acetic acid than another is expected to have 0.471 g/dm33 less citric acidQuestion 2.4 (2 marks)One of the bottles of red wines had 0.08 g/dm33 of citric acid and 0.62 g/dm33 of tartaric acid.The value of the residual for this bottle of red wine is: Answer (round your answer to 3dp)Question 2.5 (4 marks)Is the normality condition satisfied?Based on the Answer, as we Answer observe Answer, we can say that the condition of normality is Answer.Question 2.6 (1 mark)What other conditions, in addition to normality, must be satisfied for the linear model to be valid?Choose ONE option.All expected counts are at least 5, representative, paired.Independence, equal spread and linearity.Paired, linearity and independence.Sample size at least 30, representative and linearity.Question 2.7 (3 marks)The test for the slope has a p-value of 1.2949329 × 10-19Decision: As the p-value is Answer than 0.05, we find in favour of the Answer.Conclusion:Choose ONE option.There is evidence that the slope is equal to -0.471There is evidence that the slope is not equal to 0.5There is evidence that the slope is not equal to 0There is evidence that the slope is equal to 0

the main red grape used in the wines of Burgundy?

Measurements for the titration of acetic acid with NaOHEnter your measurements from lab for trails 1 - 3 with the correct number of decimal places  Trial 1Trial 2Trial 3Mass of empty flask (grams)71.39171.42572.095Mass of flask + vinegar (grams)81.00381.37982.014Volume of vinegar (L)0.010000.010000.01000Concentration of NaOH (mol/L)0.20000.20.2Initial volume of NaOH (mL)0.600.290.20Final volume of NaOH (mL)41.6541.9441.53AttendanceThis QR code needs to be scanned by your TA to confirm that you have completed data collection and cleaned up your workspace.Done online(51pts) CalculationsTitration of Acetic Acid with NaOHCalculations for the titration of acetic acid with NaOHTable view List viewComplete the calculations below for trials 1 - 3 with the correct number of decimal places  Trial 1Trial 2Trial 3Mass of empty flask (grams)71.39171.42572.095Mass of flask + vinegar (grams)81.00381.37982.014Mass of vinegar (grams)Volume of vinegar (L)0.010000.010000.01000Concentration of NaOH (mol/L)0.20000.20000.2000Initial volume of NaOH (mL)0.600.290.20Final volume of NaOH (mL)41.6541.9441.53Volume of NaOH added (mL)Volume of NaOH added (L)Moles of NaOH used (mol)Moles of acetic acid (mol)Mass of acetic acid (grams)Mass percent of acetic acid(9pts)Average mass percent(9pts)Standard Deviation

Some friends are trying to make wine in their basement. They've added yeast to a sweet grape juice mixture and have allowed the yeast to grow. After several days they find that sugar levels in the grape juice have dropped, but there's no alcohol in the mixture. The most likely explanation is that ______.the mixture needs less oxygen; yeast only produce alcohol in the absence of oxygenthe mixture needs more sugar; yeast need a lot of energy before they can begin to produce alcoholthe mixture needs more oxygen; yeast need oxygen to break down sugar to produce alcoholthe mixture needs less sugar; high sugar concentrations stimulate cellular respiration, and alcohol is not a by-product of cellular respiration

The reaction involved in this experiment is described as acetylation of Salicylic Acid. Discuss how the acetylation reaction is useful in the formation of other drug molecules from opiate