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Gelneren [198K]

3 years ago

12

Antifreeze is mixed with water in a car's cooling system because the solution what?

1 answer:

Nikolay [14]3 years ago

8 0

Antifreeze is an additive in water-based liquid to lower down the freezing point of such liquid. It is used to make use of the colligative properties of solutions specifically freezing-point depression for cold climate and boiling-point elevation to allow higher coolant temperature.

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How many atoms are in 0.00100 mole of Na?

Ymorist [56]

There rlly isn’t noooo number so what do you expect of course it would be zero. 0£

8 0

3 years ago

Read 2 more answers

When H2SO4 is added to PbI2, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If

Nookie1986 [14]

Answer:

$n_{I^-}=3.11x10^{-4}molI^-$

Explanation:

Hello,

In this case, the undergoing chemical reaction is shown below:

$H_2SO_4(aq)+PbI_2(aq)\rightarrow PbSO_4(s)+2HI(aq)$

Thus, we obtain 0.0471 g of lead (II) sulfate, the iodide ions in the original solution are computed below by stoichiometry, taking into account that into 1 mole of lead (II) iodide there are 2 moles of iodide ions:

$n_{I^-}=0.0471gPbSO_4*\frac{1molPbSO_4}{303.26gPbSO_4}*\frac{1molPbI_2}{1molPbSO_4}*\frac{2molI^-}{1molPbI_2} \\n_{I^-}=3.11x10^{-4}molI^-$

Best regards.

3 0

3 years ago

Which chamber collects blood from the lungs that is rich in oxygen?

Umnica [9.8K]

The left atrium does that

7 0

3 years ago

what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

5 0

3 years ago

What does a plant produce during photosynthesis? A. Oxygen and glucose B. Carbohydrates and carbon C. Carbon, oxygen, and hydrog

Nataliya [291]

Answer:

im pretty sure b

Explanation:

3 0

3 years ago