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3 years ago

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If a 60 kg dog is on a 15 kg sled is pushed with a force of 300 N, what will be dogs acceleration? A) 900 N B) 4 kg C) 4 m/s2 D)

0.25 m/s2

2 answers:

Mamont248 [21]3 years ago

7 0

The answer is C, because 60 divided by 15 is 4.
Hope this Helps!
Mark me Brainliest, Please

stepan [7]3 years ago

6 0

The answer is C because you have to find acceleration.

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4 0

3 years ago

Read 2 more answers

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

<u>Step 3: </u>Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]

[Ca2+]= 0.050 M [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

<u> Step 7:</u> Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0

3 years ago

when 10.00 g of phosphorus reacts with oxygen, it produces 17.77 g of a phosphorus oxide. This phosphorus oxide was found to hav

Alex_Xolod [135]

Answer:

Molecular formula = P₄O₆

Explanation:

P(s) + O₂(g)------------------------------------⇒ PₓOₙ (g)

10g (17.77-10)g 17.77g

10g 7.77g 17.77g (gramme ratio)

The molecular mass of Phosphorus (P) = 31g/mole

The molecular mass of Oxygen atom (O) = 16g/mole

Mole ratio is given by:

P : O

10/31 7.77/16

0.3226 : 0.4856 Mole ratio---------------------------- (1)

Divide (1) through by 0.3226

1 : 1.5-------------------------------------------- (2)

From (2), the empirical formula for Phosphorus oxide :

Empirical formula = P₁O₁.₅

= PO₁.₅

The molecular formula can be calculated from below:

Since the molecular formula is a multiple of the empirical formula we have

Molecular formula = (PO₁.₅)ₙ----------------------------------- (3)

Since we are given the molecular mass of the oxide formed, we have:

(PO₁.₅)ₙ = 220-----------------------------(4)

[31 + (16 x 1.5)] x n = 220

[31 + 24]n = 220

55n =220

n = 4

Substituting into (3), we have :

Molecular formula = (PO₁.₅)₄

= P₄O₆

8 0

3 years ago

The correct mathematical expression for finding the molar solubility (s) of sn(oh)2 is:

lara [203]

Answer is: Ksp = 4s³.
Balanced chemical reaction (dissociation) of strontium hydroxide:
Sr(OH)₂(s) → Sr²⁺(aq) + 2OH⁻(aq).
Ksp(Sr(OH)₂) = [Sr²⁺]·[OH⁻]².<span>
[</span>Sr²⁺] = s.<span>
[</span>OH⁻] = [Sr²⁺] = 2s<span>
Ksp = (2s)² · x = 4s³.
Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.
[</span>Sr²⁺] is equilibrium concentration of iumcations.<span>
[</span>OH⁻] is equilibrium concentration of hydroxide anions.

7 0

3 years ago

What term expresses the ratio of moles of reactants and products according to the coefficients in the balanced chemical equation

bogdanovich [222]

Answer:

A) mole ratio

Explanation:

<em>A mole ratio</em>, also known as a mole-to-mole ratio, <em>is the ratio between the amounts in moles of one reactant/product to the moles of the other reactant/product. </em>This ratio is determined considering the coefficients in a balanced chemical equation. This ratio is used in chemical problems as a conversion factor between the compounds involved in the reaction.

I hope you find this information useful and interesting! Good luck!

6 0

3 years ago