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Likurg_2 [28]
3 years ago
6

Units used to measure the rate of electron flow

Chemistry
1 answer:
AleksandrR [38]3 years ago
3 0
SI units = ammeters, their called amps
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In 1990, the northern spotted owl was listed as endangered. Because of this, millions of acres of forests in the Pacific
Gwar [14]

Answer:

the endangered spices act

Explanation:  The US Endangered Species Act (ESA) is our nation's most effective law to protect at-risk species from extinction, with a stellar success rate: 99% of species listed on it have avoided extinction.

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What are the 5 common uses of fossil fuels?
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Electricity. Coal alone provides half the electricity in the United States. ...

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Explanation:

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3 years ago
How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be usefu
In-s [12.5K]

Answer:

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.

q 1= m c (T2-T1)

q1 = 36.0 g (4.18 J/gC) (100 - 65 C)

q1 = 5267 J =5.267kJ

next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2= 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3= 36.0 g (2.01 J/gC)(115-100C)

q3 = 1085 J =1.085kJ

qnet=q1 +q2 +q3

energy required=qnet=87.75kJ

8 0
3 years ago
Baking soda and vinegar can be used as a household drain cleaner, by putting the baking soda powder into the kitchen drain say,
Mrac [35]

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Oxygren co2 and nitrogen

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7 0
2 years ago
Read 2 more answers
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
3 years ago
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