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max2010maxim [7]
3 years ago
6

Maize, man, Goat, is an example of a food chaintrue or false​

Chemistry
1 answer:
Lina20 [59]3 years ago
8 0

Answer:

False

Explanation:

The correnct chain is labeled Maize, Goat, Man.

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What size volumetric flask would you use to create a 0.50 M<br> solution using 10.00 g of CaBr2
SIZIF [17.4K]

Answer:

Explanation:

First convert the grams of Calcium Bromide to moles by using the atomic weight. Then use the formula for molarity, which is moles per liter.

CaBr2 = 199.9 g/mol

10/199.9 = 0.05 moles of CaBr2

0.5M=\frac{0.05mol}{x}

x = 0.1L or 100mL

5 0
2 years ago
1. Define a Family.
Nataliya [291]

Answer:

  1. a family is a group consisting of 2 parents and their children living together in a union.

or all desendants of a common ancestor

2. A period is a name given to the horizontal row of the periodic table.

3.

a.mg

b.k

c.Fe

d.cu

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a.Carbon

b.Chlorine

c.Gold

d.strontium

5.

a.period 1

b.period 4

c.period 5

d.period 5

6.

a.group 6

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c.group 7

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a.F

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4 0
2 years ago
List the 5 chemicals with the highest flash point. (In order, staring with the highest)
Volgvan

Answer:

Flash point of Lube Oil is around the 187°C mark, 

Flash point of Biodiesel of 130°C

Flash point of Diesel Ranging from 52° to 96°

8 0
3 years ago
I WILL MARK BRAINLIEST!Which of the following statements is true?
PtichkaEL [24]

Answer:

B- fission breaks apart nuclei; fusion puts them back together

6 0
3 years ago
Read 2 more answers
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
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