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3 years ago

C. Convert 1.05x107 milliliters to kiloliters (L = liter)

Chemistry

2 answers:

Oksana_A [137]3 years ago

5 0

1.05 into 107 millilitres is equal to 10.5 kilolitres
Hope this helps. :)

kari74 [83]3 years ago

3 0

Answer:

10.5 kl

Explanation:

$1.05*10^7ml\\=\frac{1.05*10^7}{10^3}l\ [1 ml= 1/1000 l]\\=1.05*10^4\\=10500 l\\=\frac{10500}{1000} l\ [1l=1/1000kl]\\=10.5 kl$

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78.9 + 890.43 - 21 = 9.5 x 10^2

maria [59]

948 or 9.48 x 10^2

There are two sets of rules for significant figures

• One set for addition and subtraction

• Another set for multiplication and division

You used the set for multiplication and division.

This problem involves addition and subtraction, and the rule is

The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.

Thus, we have

78.9

+890.43

-21.

= 948.33

The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.

To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.

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3 years ago

GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g

JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of $H_2$ gas and 1 mole of $Br_2$ liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

$H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ$

Divide the equation by 2.

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

The equation to show the the correct form to show the standard molar enthalpy of formation:

$\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ$

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3 years ago

What will the outcome of putting litumus paper into distilled water

Kay [80]

If it is blue litmus paper, it will remain blue. If it is red litmus paper, it will remain red. Water is neither an acid nor a base so it will not change the litmus paper.

5 0

3 years ago

2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme

mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

$2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O$

Next, we assign the oxidation numbers as follows:

$2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}$

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

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2 years ago

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Nadya [2.5K]

Plants don't need sap. They make it like Maple trees.

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