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Likurg_2 [28]
3 years ago
6

HELP WILL GIVE BRAINLIESTWhich of these elements would have the largest ionic radius?

Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

Bi3+ has 103

C4- has 170

Li+ 76

Na+102

Cl-185

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What is the pressure in a 5.00 L tank with 49.00 grams of oxygen gas at 350 K? ___atm
ziro4ka [17]

Answer:

The right answer is "8.81 atm".

Explanation:

Given:

V = 5.00 L

Mass = 4900 g

MW = 32 g/mol

T = 350 K

Now,

Number of moles will be:

n = \frac{Mass}{MW}

   =\frac{49.00}{32}

   =1.53125 \ mole

By using the ideal gas equation, we get

⇒ PV=nRT

or,

⇒    P=\frac{nRT}{V}

By substituting the values, we get

           =\frac{1.53125\times 0.0821\times 350}{5.00}

           =\frac{44.1}{5.00}

           =8.81 \ atm  

4 0
3 years ago
Use the Gizmo to estimate the age of each of the objects below. For these questions, each second in the Gizmo represents 1,000 y
ziro4ka [17]

Answer:

Check the explanation

Explanation:

AT = A0 e(-T/H)

... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.

Substituting what we know, we get...

0.71 = (1) e(-T/5730)

Solve for T...

loge(0.71) = -T/5730

T = -loge(0.71)(5730)

T = 1962 (conservatively rounded, T = 2000)

similarly for all

for aboriginal charcoal

0.28 = (1) e(-T/5730)

Solve for T...

loge(0.28) = -T/5730

T = -loge(0.28)(5730)

T = 7294 (conservatively rounded, T = 7000)

for mayan headdress

0.89 = (1) e(-T/5730)

Solve for T...

loge(0.89) = -T/5730

T = -loge(0.89)(5730)

T = 667 (conservatively rounded, T = 700)

for neanderthal

0.05 = (1) e(-T/5730)

Solve for T...

loge(0.05) = -T/5730

T = -loge(0.05)(5730)

T = 17165 (conservatively rounded, T = 17000)

7 0
3 years ago
What are these smooth, distinct layers most directly evidence of ?
Neko [114]

Answer: The layers of the rocks in one region of the parks are smooth and distinct, which are evidence of many, many years of deposition. The layers on the rocks are because of different deposition of sediments. Different sediments deposited over the rocks through wind, water and ice over the ages

Explanation:

5 0
2 years ago
Read 2 more answers
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
Calculate the molarity of a nitric acid solution if 38 ml of the solution is neutralized by 16 ml of 0.25 M barium hydroxide sol
RUDIKE [14]

Answer:

l

Explanation:

8 0
3 years ago
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