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3 years ago

A 50 kg cart has 324 kg•m/s of momentum. How fast is the cart going?

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

5 0

Answer: 6.48m/s

Explanation:

From the question given, we obtained the following:

M = 50 kg

Velocity =?

Momentum = 324 kg•m/s

Momentum = Mass x Velocity

Velocity = momentum /Mass

Velocity = 324 / 50

Velocity = 6.48m/s

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Which of the following is the largest volume? (B?)

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Yeah, it would be B (1.2 x 102 m^3) because the measurement gave it away even though other numbers were higher, however, the measurements for those were smaller in size.

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Determine whether each statement is an example of a physical change or a chemical change. Newspaper yellows in the sun. Physical

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Answer:

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2 years ago

Uses of baking powder

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3 years ago

Read 2 more answers

20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

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3 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

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3 years ago