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4 years ago

I am at the centre of the atom and contain both neutrons and protons. What am I?

2 answers:

8 0

An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral.

Burka [1]4 years ago

7 0

Answer:

Nucleus and atoms have no overall charge because each element has the same number of protons and electrons which cancel out each other.

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Write a goal of the lab or the question you tried to answer

Free_Kalibri [48]

Answer:

I need to see the lab

Explanation:

7 0

3 years ago

The temperature of a system rises by 45°C during a heating process. Express this rise in temperature in Kelvin. (Round the final

postnew [5]

Answer:

45K

Explanation:

Rise in temperature = Final - initial temperature.

temperature in K = Temperature in Celsius + 273

for Celsius; T2 -T1 =45°C

for kelvin; T2+273 -(T1+273) = ?

T2+273 -T1-273 =?

T2-T1 = ?

T2-T1 =45k

hence ΔT(K) = ΔT(°C) (temperature difference in Celsius is equal to temperature difference in kelvin)

4 0

3 years ago

Classify each phrase according to whether it applies to photophosphorylation, oxidative phosphorylation, or both

chubhunter [2.5K]

Answer:

1. Both

2. Phosphorylation

3. Both

4. Phosphorylation

5. Oxidative.

6. Both

Explanation:

Phosphorylation only occurs in chloroplast and it involves larger electrical component. Both Phosphorylation and oxidative occurs in mitochondria and it involves proton gradient. They occur in plants to produce ATP. Oxidative involves in smaller electrical component.

6 0

3 years ago

What is the number of orbit for calcium

CaHeK987 [17]

It has 20 electrons

7 0

3 years ago

Question 22

ruslelena [56]

Answer:

O2, oxygen.

Explanation:

Hello.

In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

$n_{CO_2}^{by\ CH_4}=85gCH_4*\frac{1molCH_4}{16gCH_4} *\frac{1molCO_2}{1molCH_4} =5.3molCO_2\\\\n_{CO_2}^{by\ O_2}=320gO_2*\frac{1molO_2}{32gO_2} *\frac{1molCO_2}{2molO_2} =5molCO_2$

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

Best regards.

5 0

3 years ago