A nurse is providing dietary teaching to the guardian of a preschooler who has celiac disease.
Which of the following foods should the nurse recommend including in the preschooler's diet?
Corn tortilla with black beans.
Whole wheat pasta with shrimp.
Low sodium vegetable soup with barley.
A bologna sandwich on rye bread.
The Correct Answer is A
Choice A rationale:
Preschoolers with celiac disease need to avoid gluten-containing grains such as wheat, barley, and rye. Corn tortilla with black beans is a suitable option as it does not contain gluten and provides essential nutrients.
Choice B rationale:
Whole wheat pasta contains gluten, which should be avoided by individuals with celiac disease. This option is inappropriate for the preschooler with celiac disease.
Choice C rationale:
Low sodium vegetable soup with barley contains gluten, which is not suitable for a child with celiac disease. Barley is a gluten-containing grain and should be avoided.
Choice D rationale:
Rye bread contains gluten and is not appropriate for a preschooler with celiac disease. This option is not suitable for the child's dietary needs.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is C
Explanation
Choice A rationale:
Limiting fluid intake to 1 liter per day can lead to dehydration and other health complications. It is important for the client to maintain adequate hydration, especially postpartum. This option is incorrect and potentially harmful.
Choice B rationale:
Manual expression of milk can help relieve engorgement without stimulating further milk production. This method allows the client to express milk as needed. However, it can be done even before engorgment occurs
Choice C rationale:
Wearing a snug-fitting bra can provide support and comfort.
Choice D rationale:
Applying moist heat to the breasts can stimulate milk production and relieve engorgement. However, in this case, the client wants to suppress lactation. Therefore, this option is not appropriate and may have the opposite effect of increasing milk production.
Correct Answer is ["18"]
Explanation
The client weighs 198 lb, which is equivalent to (198 ÷ 2.2 = 90kg.
Therefore, the amount of mannitol for the test dose is 0.2 g/kg x 90 kg = 18 g. The nurse should administer 18 g of mannitol IV bolus over 5 min as a test dose to the client who has severe oliguria.
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