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Nadusha1986 [10]
3 years ago
10

Help please?? I’m confused

Chemistry
1 answer:
Nady [450]3 years ago
3 0
It’s c. O stands for oxygen and h stands for hydrogen. They both do play in the roll
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Ribosome are found ____
evablogger [386]
It's found inside of a cell attached to the cytoplasm and attached to the endoplasmic reticulum. hope this helps
6 0
3 years ago
Question 2 (1 point)
Vikki [24]

Answer:D

Explanation:

Jus did this question & got it right

4 0
3 years ago
A piece of iron (C=0.449 J/g°C) and a piece of gold (C=0.128 J/g°C) have identical masses. If the iron has an initial temperatur
spin [16.1K]

Answer:

The correct answer is B. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K

Explanation:

Iron is hotter and gold is colder, therefore, according to laws of thermodynamics, iron will lose heat to gold until they are at the same temperature.

The specific heat capacity of iron(0.449) is over three times that of gold(0.128). Since masses are equal, this means that each time iron's temperature drops by one degree, the energy released it releases makes gold's temperature increase by more than 3 degrees. So gold's temperature will be climbing much faster than iron's is falling. Meaning they will meet closer to the initial temperature of iron than that of gold

4 0
3 years ago
Someone help me please.. I will mark as brainliest I promise...​
Kobotan [32]

Answer:

(a) proton

(b) neutron

(c) electron

particles in nucleus are proton and neutron.

atom is electrically neutral because no.of proton= no.of electron=6

5 0
2 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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