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garri49 [273]
2 years ago
11

How many electrons are in a neutral carbon-14 atom?.

Chemistry
1 answer:
Alla [95]2 years ago
7 0

Answer:

There are six electrons in a carbon-14.

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What color do acids turn litmus paper?
Ivanshal [37]

Answer:

red

Explanation:

........pls brainliest

8 0
2 years ago
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Convert the following masses into kilograms:
beks73 [17]

Answer:

(a): 2,300 kilograms

(b): 0.005 kilograms

(c): 2.3 × 10^-5 kilograms

(d): 155 kilograms

Explanation:

Formulas:

(a); divide the mass value by 1000

(b); divide the mass value by 1e+6

(c); divide the mass value by 1e+9

(d); multiply the mass value by 1000

5 0
2 years ago
Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W
alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0
3 years ago
Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen
MaRussiya [10]

Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}}  *\frac{18.02gH_2O}{1molH_2O}

Therefore, the resulting mass of water is:

m_{H_2O}=11.3gH_2O

Best regards!

4 0
2 years ago
In the combustion of ethane, what is/are the reactant(s)?
aleksley [76]
Reactant are those that combine or reacts to give products !!

so in combustion of ethane ; ethane and o2 are reactants so ... your answer is B !!
5 0
3 years ago
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