What is the slope of the line segment<br><br>-3<br>-1/3<br>1/3<br>3

Which information should be added to the chart in order find out who ran a greater distance?

Arisa [49]

Answer:

This question is incomplete, the complete question is:

Nancy and Hiyang are training for a race. They entered some of their training notes in a chart. Which information should be added to the chart in order find out who ran a greater distance?

The answer is C). the units used to measure distance each day

Explanation:

According to the question, Nancy and Hiyang are training for a race that involves them recording the distance they ran in a chart. Distance, as a quantity, is measured using different S.I units like metres, kilometers, miles, centimeters, etc.

However, in order to accurately discover whether Nancy or Hiyang ran a greater distance as recorded in their chart, the units used to measure distance each day must be included. This is because the unit of a quantity determines how big or small it is in comparison to another. For example, 20metres is not the same as 20centimetres.

If the unit they used in measuring their distance is not included, it will be impossible to tell what is being measured, talkless of who ran a greater distance

6 0

3 years ago

Select the conjugate acid-base pair(s). a) HI, I b) HCHO2, SO4^2- c) CO3^2-, HCI d) PO4^3-, HPO4^2-

Semenov [28]

Answer:

d) PO4^3-, HPO4^2-

Explanation:

Basically, an acid and a base which differs only by the presence or absence of proton (hydrogen ion) are called a conjugate acid-base pair.

a) HI, I

This is incorrect. For the acid, HI the conjugate base is I⁻ ion.

b) HCHO2, SO4^2-

This is incorrect, there's no relationship between both entities.

c) CO3^2-, HCI

This is incorrect, there's no relationship between both entities.

d) PO4^3-, HPO4^2-

This is correct. The difference between both entities is the Hydrogen ion. This is the conjugate acid-base pair

6 0

3 years ago

Which elements did the Curies discover?

Scilla [17]

Answer:

B on edge

Explanation:

6 0

2 years ago

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

Simora [160]

Answer:- Density is 10.5 grams per mL and the substance is Silver(Ag).

Solution:- Mass of jewelry is given as 132.6 g. Initial volume of water in the graduated cylinder is 48.6 mL and the volume becomes 61.2 mL when this jewelry is submerged in the cylinder.

So, the volume of the jewelry = 61.2 mL - 48.6mL = 12.6 mL

We know that, $density=\frac{mass}{volume}$

Let's plug in the values in the above formula to calculate the density:

$density=\frac{132.6g}{12.6mL}$

$density=10.5\frac{g}{mL}$

So, the density of the jewelry is $10.5\frac{g}{mL}$ .

The substance from which the jewelry is made could easily be identified by comparing the calculated density with the density values given for different substances in the density table.

Looking at the density table, 10.5 grams per mL is the density for silver. So, the jewelry is made up from Silver.

4 0

3 years ago

A compound has a percent composition of 81.71% C and 18.29% H. What is the empirical formula of this compound?

SCORPION-xisa [38]

Considering the definition of empirical formula, the empirical formula is C₃H₈.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, in first place you know the percent composition:

C: 81.71 %
H: 18.29%

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have 81.71 grams of carbon and 18.29 grams of hydrogen H.

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

C: $\frac{81.71 grams}{12\frac{grams}{mole} }$ = 6.81 moles
H: $\frac{18.29 grams}{1\frac{grams}{mole} }$ = 18.29 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

C: $\frac{6.81 moles}{6.81 moles}$ = 1
H: $\frac{18.29 moles}{6.81 moles}$ = 2.68 ≅ $\frac{8}{3}$