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Sedbober [7]
3 years ago
12

URGENT i need this fast

Chemistry
1 answer:
erastovalidia [21]3 years ago
5 0
Less than 250,000 joules
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several kinds of bears are found on earth. most bears are brown or black, but one type of bear, the polar bear is white. What pr
LekaFEV [45]

Explanation:

they are white so they can camouflage, but actually the polar bears skin is black and it's hair is hollow

4 0
3 years ago
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What happens when you put sugar into vinegar
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Nothing. Many salad dressings are a mixture of sugar and vinegar.
5 0
3 years ago
The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p
almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen               40300 g

Carbon                11160 g

Hydrogen            6200 g

Nitrogen              1860 g

Calcium                992 g

Phosphorus         744 g

Other elements   744 g

It is possible to check it adding all the components, the total is 62000 g.  

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

                            Percentage ( %)

Oxygen               65.0

Carbon               18.0

Hydrogen            10.0

Nitrogen              3.0

Calcium               1.6

Phosphorus         1.2

Other elements   1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

                           Percentage ( %)      Decimal number

Oxygen               65.0                         65/100 = 0.65

Carbon               18.0                           18/100 = 0.18

Hydrogen            10.0                         10/100 = 0.10

Nitrogen              3.0                           3/100 = 0.03

Calcium               1.6                           1.6/100 = 0.016

Phosphorus         1.2                           1.2/100 = 0.012

Other elements   1.2                           1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

                           Percentage ( %)      Decimal number      Mass in grams (g)        

Oxygen               65.0                         65/100 = 0.65         0.65*62000=40300

Carbon               18.0                           18/100 = 0.18         0.18*62000=  11160

Hydrogen            10.0                         10/100 = 0.10         0.10*62000=   6200

Nitrogen              3.0                           3/100 = 0.03          0.03*62000= 1860

Calcium               1.6                           1.6/100 = 0.016        0.016*62000= 992

Phosphorus         1.2                           1.2/100 = 0.012        0.012*62000=744

Other elements   1.2                           1.2/100 = 0.012        0.012*62000=744  

5 0
3 years ago
What mass of hydrogen sulfide, H2S, will completely react with 2.00 moles of silver nitrate, AgNO3?
hodyreva [135]

Answer:

34g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H2S + 2AgNO3 —> 2HNO3 + Ag2S

Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.

This is illustrated below:

From the balanced equation above,

We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.

Finally, we shall convert 1 mole of H2S to grams. This is shown below:

Number of mole H2S = 1 mole

Molar mass of H2S = (2x1) + 32 = 34g/mol

Mass = number of mole x molar Mass

Mass of H2S = 1 x 34

Mass of H2S = 34g

Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.

6 0
3 years ago
A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
givi [52]

Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

V_{V^{+3}} = 59.0 \ mL

V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
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