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IrinaVladis [17]

3 years ago

7

Your sick puppy is diagnosed with roundworms. They grow inside of the puppy. What type of symbiotic relationship do the puppy an

d the roundworms have?
A. Cooperative
B. Commensalism
C. Mutualism
D. parasitism

1 answer:

Ratling [72]3 years ago

4 0

D. Parasitism , the roundworm benefits from the relationship, but the puppy is harmed

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I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w

trasher [3.6K]

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s] + 2 HCl [aq] → BeCl 2(aq] + H 2(g] ↑ Be (s] + 2 HCl [aq] → BeCl 2(aq] + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get x 100

=3.5 g/0.45 g 100

= 777.78 %

The percentage of yield was 777.78%

5 0

3 years ago

Fire extinguishers that spray carbon dioxide on the fire, work very effectively because it forms a blanket around the burning ma

Arte-miy333 [17]

Answer:

I think it will option B it will retain enough heat

8 0

3 years ago

poizon [28]

So, what's your question..?. You just gave options..

8 0

3 years ago

Read 2 more answers

Silicon carbide (SiC) is an important ceramic material made by reacting sand (silicon dioxide, SiO2) with powdered carbon at a h

makkiz [27]

Answer:

Percent yield of SiC is 77.0%.

Explanation:

Balanced reaction: $SiO_{2}+3C\rightarrow SiC+2CO$

Molar mass of SiC = 40.11 g/mol

Molar mass of $SiO_{2}$ = 60.08 g/mol

So, 100.0 kg of $SiO_{2}$ = $\frac{100.0\times 10^{3}}{60.08}$ moles of $SiO_{2}$ = 1664 moles of $SiO_{2}$

According to balanced equation, 1 mol of $SiO_{2}$ produces 1 mol of SiC

Therefore, 1664 moles of $SiO_{2}$ produce 1664 moles of SiC

Mass of 1664 moles of SiC = $(1664\times 40.11)g$ = 66743g = 66.74 kg (4 sig. fig.)

Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)] $\times 100$ %

= $\frac{51.4kg}{66.74kg}\times 100$ %

= 77.0%

4 0

3 years ago

g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint

diamong [38]

Answer:

The value is $L = 0.985 \pm 0.00801 \ g$

Explanation:

From the question we are told that

The molar mass of $CaCO_3$ is $MW = 100.085 \ g/mol$

The total mass is $m_g = 12.3 \ g$

The uncertainty of the total mass is $\Delta g = 0.1$

Generally the molar weight of calcium is $M_c = 40 g/mol$

The percentage of calcium in calcite is mathematically represented as

$C = \frac{40.07}{100.085} * 100$

$C = 40.03 \%$

Generally the mass of each sample is mathematically represented as

$m= \frac{m_g}{5}$

$m= \frac{12.3}{5}$

$m= 2.46 \ g$

Generally mass of calcium present in a single sample is mathematically represented as

$m_c = 2.46 * \frac{40.04}{100}$

$m_c = 0.985 \ g$

The uncertainty of mass of a single sample is mathematically represented as

$k = \frac{\Delta g }{5}$

$k = \frac{0.1 }{5}$

$k = 0.02\ g$

The uncertainty of mass of calcium in a single sample is mathematically represent

$G = \frac{0.02 * 40.04}{ 100}$

$G = 0.00801 \ g$

Generally the average mass of calcium in each sample is

$L = 0.985 \pm 0.00801$

6 0

3 years ago