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4 years ago

Calculate volume of carbon dioxide gas if the mass is 88.0 g at 20.0°C and 950mmHg

Chemistry

1 answer:

aivan3 [116]4 years ago

7 0

Answer:

Tam tam fdjknf

Explanation:

ttgtsrttrg

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Two charged particles repel each other with a force of F.

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Answer:

it would increase to twice what it was

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3 years ago

A team of scientists have identifyed a group of fish in the San Franscisco Bay that have never been seen in this location before

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4 years ago

Read 2 more answers

The steps required to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25x10^-2 M. Please

Yanka [14]

Answer:

In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.

Explanation:

Concentration: 1.25 x 10⁻² M

1,25 x 10⁻² mol FeCl₃ ___ 1000 mL

x ___ 200.0 mL

x = 2.5 x 10⁻³ mol FeCl₃

Mass of FeCl₃:

1 mol FeCl₃ _____________ 162.2 g

2.5 x 10⁻³ mol FeCl₃ _______ y

y = 0.4055 g FeCl₃

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4 years ago

Write 0.000 000 000 402 000 in scientific notation.

Rina8888 [55]

Answer:

0.000000000402000 in scientific notation is 4.02000 × 10-10

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3 years ago

A. Which reactant is the limiting reagent?

Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. $m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2$

c. $m_{Br_2}^{leftover}=6.6g$

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

$Zn+Br_2\rightarrow ZnBr_2$

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

$n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2$

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

$m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2$

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

$m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2$

Thus, the leftover of bromine is:

$m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g$

Best regards!

8 0

3 years ago