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Allisa [31]
3 years ago
7

Please help me with this homework

Chemistry
1 answer:
Stolb23 [73]3 years ago
6 0
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A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s.
Sedaia [141]

Answer:

F centripetal force (tension) = 275.9  N

Explanation:

Given data:

Mass = 1.50 kg

Radius = 0.520 m

Velocity of ball = 9.78 m/s

Tension = ?

Solution:

F centripetal force (tension) =  m.v² / R

F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m

F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m

F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m

F centripetal force (tension) = 275.9  N

7 0
3 years ago
Hydrogen ion reacts with zinc to produce_____ gas?
WINSTONCH [101]

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

Hydrogen ion reacts with zinc to produce\:\pmb{\underline{\red{\sf{Zinc \:hydride    }}}}\:gas

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

6 0
3 years ago
How does a chemical indicator work ?
sukhopar [10]
PH indicators detect the presence of H+ and OH-. They do this by reacting with H+ and OH-: they are themselves weak acids and bases. If an indicator is a weak acid and is coloured and its conjugate base has a different colour, deprotonation causes a colour change.
5 0
4 years ago
Read 2 more answers
A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu
klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}

<u>T₂ = 19.95°C</u>

6 0
3 years ago
Zn(s)+2HCl(aq)-&gt;ZnCl2(aq)+H2(g)
kobusy [5.1K]
To answer this we need to assume that at STP 1 mol of a substance is equal to 22.4 L. We use this data for the calculations. 
Statement 1
2 mol HCl (1 mol H2 / 2 mol HCl) = 1 mol H2 thus at STP 22.4 L H2
<em>This statement is false.</em>
Statement 2
1 L Zn (1 mol Zn / 22.4 L Zn) (1 mol H2 / 1 mol Zn) (22.4 L H2 / 1 mol H2) = 1L H2
<em>This statement is true.</em>
Statement 3
65.39 g Zn (1 mol Zn / 65.39 g Zn) (1 mol H2 / 1 mol Zn) = 1 mol H2
<em>This statement is true.</em>
Statement 4
1L HCl (1 mol HCl / 22.4 L HCl)(1 mol H2 / 2 mol HCl) = 0.022 mol H2
<em>This statement is false.</em>
7 0
3 years ago
Read 2 more answers
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