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1 year ago

5

Danny sails a boat downstream. The wind pushes the boat along at 21 km/hr. The current runs downstream at 15 km/hr. What is the

actual velocity of the boat?
A) 6.0 km/hr, downstream
B) 1.4 km/hr, downstream
C) 310 km/hr, downstream
D) 36 km/hr, downstream

2 answers:

deff fn [24]1 year ago

7 0

Answer: D) 36.0 km/hr, downstream

Explanation:

For downstream motion of the boat, the actual velocity of the boat is the sum of velocity of the water current and the velocity of the boat due to pushed by wind.

Velocity of water current, v = 15 km/h

Velocity of the boat going downstream, u = 21 km/h

Actual velocity of the boat = v'

v' = v + u

⇒v' = 15 km/h + 21 km/h

⇒u = 21 km/h +15 km/h = 36.0 km/h downstream

Thus, the correct answer is option D.

natali 33 [55]1 year ago

4 0

D. 36 km/hr, downstream

you add the boats speed of 21kph to the rivers speed of 15kph to get your answer

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The elements chlorine and iodine have similar chemical properties because they _____. Group of answer choices have the same numb

worty [1.4K]

Answer:

Have the same number of electrons in their outer energy levels

Explanation:

Elements in the same group have similar chemical properties because they have the same number of valence electron(s) in their outermost shell.

Chlorine and Iodine have similar chemical properties because they have the same number of valence electron in their outermost shell. This can be seen from their electronic configuration as shown below:

Cl (17) => 1s² 2s²2p⁶ 3s²3p⁵

I (53) => [Kr] 4d¹⁰ 5s²5p⁵

From the above illustration:

Outer shell of Cl (3s²3p⁵) = 2 + 5 = 7 electrons

Outer shell of I (5s²5p⁵) = 2 + 5 = 7 electrons

Since they have the same number of valence electrons, therefore, they will have similar chemical properties.

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1 year ago

Using the following thermochemical equation, determine the amount of heat produced from the combustion of 24.3 g benzene (C6H6).

Anna007 [38]

Answer:

ΔH = -976.5 kJ

Explanation:

For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.

The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).

So, 24.3 g of benzene has :

n = mass/ molar mass

n = 24.3/78.11

n = 0.311 moles

2 moles ------------ -6278 kJ

0.311 moles ----------- x

By a simple direct three rule:

2x = -1953.08

x = -976.5 kJ

3 0

1 year ago

Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?

muminat

1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>

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2 years ago

Read 2 more answers

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

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2 years ago