What type of reproduction occurs in members of the bacteria kingdom?

The dessert can be harsh environment with extreme heat during the day and cold during the night also there is very little rainfa

Sphinxa [80]

Answer:

True

Explanation:

The desert refers to a region of arid land which is characterized by extreme temperatures, extreme dryness, low amount of precipitation and generally harsh living conditions. Because of these harsh conditions, they have been tagged with various names ranging from 'Death Valley' to 'the place from where there is no return' etc.

Every desert is made up of 2 components: the <u>biotic (living) component</u> and the <u>abiotic (non-living) component</u>. The biotic (living) component consists of the plants and animals that have adapted to these harsh living conditions e.g. Cactus or Cacti, Holly plants, Camels, Lizards, Snakes etc. The abiotic (non-living) component consists of climate (subtropical deserts which are extremely cold or temperate deserts which are extremely hot), location, precipitation/rainfall

8 0

3 years ago

A student prepares a weak acid solution by dissolving 0.2000 g HZ to give 100. mL solution. The titration requires 33.5 mL of 0.

nexus9112 [7]

Answer:

a)M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88 g/mol

b) if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94 hence the molar mass will be too low

Explanation:

0.2000 gHZ gives 100ml acid solution

33.5 ml of 0.1025 M NaOH is required to prepare it

the moles = mass / molar mass

mass = 0.200 gHZ

moles = 0.0335*100 * 0.1025 = 0.034

therefore molar mass = mass / moles

M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88

if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94 hence the molar mass will be too low

3 0

3 years ago

Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.

Sophie [7]

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

$y=-3.5 x + 25\\\\y=-0.52 x + 2$

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

$-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72$

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

$y=-3.5 (7.72) + 25\\\\y = 1.84$

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

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