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solong [7]
3 years ago
13

Round 1047.78 to three sig figs 105 100 1030

Chemistry
1 answer:
Komok [63]3 years ago
7 0

Answer:

<em>1</em><em>0</em><em>4</em><em>7</em><em>.</em><em>7</em><em>8</em>

<em>nearest</em><em> </em><em>tens</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em>4</em><em>7</em><em>.</em><em>7</em><em>0</em>

<em>near</em><em>est</em><em> </em><em>thousandth</em><em> </em><em>is</em><em> </em><em>1</em><em>0</em><em>5</em><em>0</em><em>.</em><em>7</em><em>8</em>

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Which material is likely to slow the<br> flow of electric charges the most? Explain.
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5 0
3 years ago
4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration
anzhelika [568]

Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

<h3>Definition of percent composition </h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

<h3>Percentage Composition in this case</h3>

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100

<u><em>percentage composition of potassium iodide= 0.017%</em></u>

Finally, the percent composition of potassium iodide in this sample is 0.017%.

Learn more about percent composition:

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2 years ago
if the caffeine concentration in a particular brand of soda is 2.87 mg/oz, drinking how many cans of soda would be lethal? assum
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3 years ago
The enthalpy change for converting 1.00 mol of ice at -25.0 °c to water at 70.0 °c is __________ kj. the specific heats of ice
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Answer is: 6,16 kJ.
1) changing  temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
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m(H₂O) = 1mol · 18 g/mol = 18 g.
C - <span>specific heat of ice.
</span>2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C 
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C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.

8 0
3 years ago
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