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frosja888 [35]
3 years ago
10

Resistance to electricity is measured in___________ Watts Amps Volts Ohms

Chemistry
2 answers:
Dmitry [639]3 years ago
8 0
It's D. Ohms!! I remember this from my old science class!!! Hope this helps!!!! :)
Umnica [9.8K]3 years ago
4 0
The resistance to electricity is measured in Ohms
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There are four hundred participants in the local Thanksgiving 5K race. If 51% of the participants are female, what is the ratio
Paha777 [63]

Answer:

51:49 is the ratio of female participants to male participants.

Explanation:

Percentage of females in the race = 51 %

Percentage of males in the race = 100% - 51 % = 49%

Total participants in race = 400

Number of women participants = 51% of 400=\frac{51}{100}\times 400=204

Number of men participants  = 49% of 400=\frac{49}{100}\times 400=196

Ratio of female participants to male participants :

\frac{204}{196}=\frac{51}{49}=51:49

5 0
4 years ago
Pls answer this question and have a AMAGING day!! :))
ladessa [460]

Answer:

B

Explanation:

because it is being cooled down

hoped this helps

3 0
3 years ago
Classify each reaction as combination, decomposition, single replacement, double replacement, or combustion. c2h4(g) + 3o2(g) →2
soldier1979 [14.2K]
If combined with oxygen. the reaction is usually combustion. in this case, it is the combustion of ethene
5 0
3 years ago
Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
What is the mass of oxygen in 10.0 grams of water
Afina-wow [57]

The answer is 16.00 amu.

7 0
4 years ago
Read 2 more answers
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