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mylen [45]
2 years 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
Chemistry
2 answers:
deff fn [24]2 years 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]2 years 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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Combining aqueous solutions of bai2 and na2so4 affords a precipitate of baso4. Which ions are spectator ions in the reaction?
Marina86 [1]

Answer:

I⁻ (aq) and Na⁺ (aq)

Explanation:

We have the chemical reaction:

BaI₂ + Na₂SO₄ → BaSO₄ + 2 NaI

However if you want to determine the spectator ions you need  to write the states of compounds:

(aq) - ions dissolved in water

(s) - solid

Ba²⁺ (aq) + 2 I⁻ (aq) + 2 Na⁺ (aq) + SO₄²⁻ (aq) → BaSO₄ (s) + 2 Na⁺ (aq) + 2 I⁻ (aq)

The ions which does not change the state and remains dissolved in the solution are spectator ions. For our chemical reaction we have the following spectator ions:

I⁻ (aq) and Na⁺ (aq)

3 0
2 years ago
Two different compounds are obtained by combining nitrogen with oxygen. The first compound results from combining 46.7 gg of NN
mina [271]

Answer:

4.02

Explanation:

The mass ratios will be given by dividing the mass of O₂ into the mass of N₂.

So lets do our calculations:

First Compound:

53.3 g O₂ / 46.7 g N₂ =  1.14

Second Compound:

82.0 g O₂ / 17.9 g N₂ = 4.58

Ratio = 4.58 / 1.14 =  4.02

This result for all practical purposes is a whole number, and it is telling us that there are 4 times as many oxygen atoms in the second coumpound as in the first compound. This is so because the ratio we just calculated is also the ratio in mol atoms:

Ratio = [ mass O₂ / MW O2/ mass N₂/ MW N₂] 2nd compound  /   [mass O₂ / MW O2/ mass N₂/ MW N₂  !st compound]

and the molecular weights cancel each other.

The only N and O compounds that follow this ratio are N₂O₄ and N₂O, and this question could be made in a multiple choice to match  formulas.

3 0
2 years ago
Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.
jolli1 [7]

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, 0.5^{n}

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

8 0
2 years ago
Read 2 more answers
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
Based on the products obtained, rank the functional groups (acetamido, amino, and methoxy) in order of increasing ability to act
I am Lyosha [343]

Answer:

Amino >Methoxy > Acetamido

Explanation:

Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.

The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.

The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.

Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.

Therefore, decreasing order will be as follows:

Amino >Methoxy > Acetamido

7 0
2 years ago
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