Is Ag NO2 iconic bond or convalent bond

Pay if forward movie questions worksheet

stellarik [79]

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2:

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5 0

4 years ago

Help please :)<br> Describe two characteristics that protists have that make them eukaryotes.

V125BC [204]

Answer:

Animal? Bacteria? Plant? Fungi? What do these figures represent?

2

None of the above! These organisms may be single-celled like bacteria, and they may look like a fungus. They also may hunt for food like an animal or photosynthesize like a plant. And, yet, they do not fit into any of these groups. These organisms are protists!

What are Protists?

3

Protists are eukaryotes, which means their cells have a nucleus and other membrane-bound organelles. Most protists are single-celled. Other than these features, they have very little in common. You can think about protists as all eukaryotic organisms that are neither animals, nor plants, nor fungi.

4

Although Ernst Haeckel set up the Kingdom Protista in 1866, this kingdom was not accepted by the scientific world until the 1960s. These unique organisms can be so different from each other that sometimes Protista is called the “junk drawer" kingdom. Just like a junk drawer, which contains items that don't fit into any other category, this kingdom contains the eukaryotes that cannot be put into any other kingdom. Therefore, protists can seem very different from one another.

Explanation:

Hope it helps, some how.

7 0

3 years ago

A 346.9 ml sample of carbon dioxide was heated to 381 k. if the volume of the carbon dioxide sample at 381 k is 596.2 ml, what w

kolbaska11 [484]

The temperature at 346.9 ml is calculated using Charles law formula

that is V1/T1= V2/T2

T1= 381 K
V1= 596.2 ml
V2 = 346.9 ml
T2=?
by making T2 the subject of the formula T2= T1V2/V1

T2 = 381 x 346.9/596.2 = 221.69 K

5 0

3 years ago

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 42. g of butane is m

taurus [48]

Answer:

127 grams of carbon dioxide

Explanation:

We need to determine the chemical equation first. Butane has a chemical formula of $C_4H_{10}$ , oxygen is $O_2$ , carbon dioxide is $CO_2$ , and water is $H_2O$ . The reactants are butane and oxygen and the products are carbon dioxide and water. So we write:

$C_4H_{10}+O_2$ ⇒ $CO_2+H_2O$

But remember! We need to balance this. Currently, there are 4 carbon atoms (C), 10 hydrogen atoms (H), and 2 oxygen atoms (O) on the left, while there are 1 carbon atom (C), 2 hydrogen atoms (H), and 3 oxygen atoms (O) on the right. Let's place a coefficient of 4 in front of the carbon dioxide and a coefficient of 5 on the water, so that we have equal numbers of carbon and hydrogen atoms on each side:

$C_4H_{10}+O_2$ ⇒ $4CO_2+5H_2O$

However, we need to ensure that there are equal numbers of O atoms, as well. On the left, we have 2 and on the right we have 13, so let's put a coefficient of 6.5 on the oxygen:

$C_4H_{10}+6.5O_2$ ⇒ $4CO_2+5H_2O$

Finally, multiply everything by 2 to get whole number coefficients:

$2C_4H_{10}+13O_2$ ⇒ $8CO_2+10H_2O$

Ah, now we can actually get to the problem!

We need to determine the limiting reactant, so let's convert the 42 g of butane and 150 g of oxygen into moles of any product, say, carbon dioxide. To convert to moles, we need to find the molar mass of each compound.

The molar mass of butane is 4 * 12.01 + 10 * 1.01 = 58.14 g/mol, while the molar mass of oxygen is 2 * 16 = 32 g/mol. We can now set up the equations:

$42 gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}} *\frac{8molCO_2}{2molC_4H_{10}} =2.8896molCO_2$

$150 gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{13molO_2} =2.8846molCO_2$

Clearly, we see that 2.8846 < 2.8896, which means that oxygen is the limiting reactant. In other words, the most products can be made when the oxygen is all used up.

Now let's finally convert moles of carbon dioxide into grams by multiplying by its molar mass, which is 12.01 + 2 * 16 = 44.01 g/mol:

$2.8846molCO_2*\frac{44.01gCO_2}{1molCO_2} =127gCO_2$

Notice that we have 3 significant figures because we had 3 significant figures at the start with 150. grams of oxygen.

<em>~ an aesthetics lover</em>

4 0

3 years ago

Urea is a common fertilizer with the formula CO(NH2)2 that is sometimes used as a chemical de-icer for icy road surfaces in the

Lunna [17]

Answer:

The answer to your question is molality = 0.61

Explanation:

Freezing point is the temperature at which a liquid turns into a solid if a solute is added to a solution, the freezing point changes.

Data

Kf = 1.86 °C/m

molality = ?

ΔTc = 1.13°C

Formula

ΔTc = kcm

Solve for m

m = ΔTc/kc

Substitution

m = 1.13 / 1.86

Simplification and result

m = 0.61

4 0

4 years ago

Is Ag NO2￼￼ iconic bond or convalent bond

Is Ag NO2 iconic bond or convalent bond